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Integration in the Plane

Module by: Lawrence Baggett. E-mail the author

Summary: A selection of theorems about integration in the plane. Theorems from other modules from the same author are referenced, and practice exercises are included.

Let SS be a closed geometric set in the plane. If ff is a real-valued function on S,S, we would like to define what it means for ff to be “integrable” and then what the “integral” of ff is. To do this, we will simply mimic our development for integration of functions on a closed interval [a,b].[a,b].

So, what should be a “step function” in this context? That is, what should is a “partition” of SS be in this context? Presumably a step function is going to be a function that is constant on the “elements” of a partition. Our idea is to replace the subintervals determined by a partition of the interval [a,b][a,b] by geometric subsets of the geometric set S.S.

Definition 1:

The overlap of two geometric sets S1S1 and S2S2 is defined to be the interior (S1S2)0(S1S2)0 of their intersection. S1S1 and S2S2 are called nonoverlapping if this overlap (S1S2)0(S1S2)0 is the empty set.

Definition 2:

A partition of a closed geometric set SS in R2R2 is a finite collection {S1,S2,...,Sn}{S1,S2,...,Sn} of nonoverlapping closed geometric sets for which i=1nSi=S;i=1nSi=S; i.e., the union of the SiSi's is all of the geometric set S.S.

The open subsets {Si0}{Si0} are called the elements of the partition.

A step function on the closed geometric set SS is a real-valued function hh on SS for which there exists a partition P={Si}P={Si} of SS such that h(z)=aih(z)=ai for all zSi0;zSi0; i.e., hh is constant on each element of the partition P.P.

REMARK One example of a partition of a geometric set, though not at all the most general kind, is the following. Suppose the geometric set SS is determined by the interval [a,b][a,b] and the two bounding functions uu and l.l. Let {x0<x1<...<xn}{x0<x1<...<xn} be a partition of the interval [a,b].[a,b]. We make a partition {Si}{Si} of SS by constructing vertical lines at the points xixi from l(xi)l(xi) to u(xi).u(xi). Then SiSi is the geometric set determined by the interval [xi-1,xi][xi-1,xi] and the two bounding functions uiui and lili that are the restrictions of uu and ll to the interval [xi-1,xi].[xi-1,xi].

A step function is constant on the open geometric sets that form the elements of some partition. We say nothing about the values of hh on the “boundaries” of these geometric sets. For a step function hh on an interval [a,b],[a,b], we do not worry about the finitely many values of hh at the endpoints of the subintervals. However, in the plane, we are ignoring the values on the boundaries, which are infinite sets. As a consequence, a step function on a geometric set may very well have an infinite range, and may not even be a bounded function, unlike the case for a step function on an interval. The idea is that the boundaries of geometric sets are “negligible” sets as far as area is concerned, so that the values of a function on these boundaries shouldn't affect the integral (average value) of the function.

Before continuing our development of the integral of functions in the plane, we digress to present an analog of (Reference) to functions that are continuous on a closed geometric set.

Theorem 1

Let ff be a continuous real-valued function whose domain is a closed geometric set S.S. Then there exists a sequence {hn}{hn} of step functions on SS that converges uniformly to f.f.

Proof

As in the proof of (Reference), we use the fact that a continuous function on a compact set is uniformly continuous.

For each positive integer n,n, let δnδn be a positive number satisfying |f(z)-f(w)|<1/n|f(z)-f(w)|<1/n if |z-w|<δn.|z-w|<δn. Such a δnδn exists by the uniform continuity of ff on S.S. Because SS is compact, it is bounded, and we let R=[a,b]×[c,d]R=[a,b]×[c,d] be a closed rectangle that contains S.S. We construct a partition {Sin}{Sin} of SS as follows. In a checkerboard fashion, we write RR as the union RinRin of small, closed rectangles satisfying

  1. If zz and ww are in Rin,Rin, then |z-w|<δn.|z-w|<δn. (The rectangles are that small.)
  2.  Rin0Rjn0=.Rin0Rjn0=. (The interiors of these small rectangles are disjoint.)

Now define Sin=SRin.Sin=SRin. Then Sin0Sjn0=,Sin0Sjn0=, and S=Sin.S=Sin. Hence, {Sin}{Sin} is a partition of S.S.

For each i,i, choose a point zinzin in Sin,Sin, and set ain=f(zin).ain=f(zin). We define a step function hnhn as follows: If zz belongs to one (and of course only one) of the open geometric sets Sin0,Sin0, set hn(z)=ain.hn(z)=ain. And, if zz does not belong to any of the open geometric sets Sin0,Sin0, set hn(z)=f(z).hn(z)=f(z). It follows immediately that hnhn is a step function.

Now, we claim that |f(z)-hn(z)|<1/n|f(z)-hn(z)|<1/n for all zS.zS. For any zz in one of the Sin0Sin0's, we have

| f ( z ) - h n ( z ) | = | f z ) - a i n | = | f ( z ) - f ( z I n ) | < 1 / n | f ( z ) - h n ( z ) | = | f z ) - a i n | = | f ( z ) - f ( z I n ) | < 1 / n
(1)

because |z-zin|<δn.|z-zin|<δn. And, for any zz not in any of the Sin0Sin0's, f(z)-hn(z)=0.f(z)-hn(z)=0. So, we have defined a sequence {hn}{hn} of step functions on S,S, and the sequence {hn}{hn} converges uniformly to ff by (Reference).

What follows now should be expected. We will define the integral of a step function hh over a geometric set SS by

S h = i = 1 n a i × A ( S i ) . S h = i = 1 n a i × A ( S i ) .
(2)

We will define a function ff on SS to be integrable if it is the uniform limit of a sequence {hn}{hn} of step functions, and we will then define the integral of ff by

S f = lim S h n . S f = lim S h n .
(3)

Everything should work out nicely. Of course, we have to check the same two consistency questions we had for the definition of the integral on [a,b],[a,b], i.e., the analogs of (Reference) and (Reference).

Theorem 2

Let SS be a closed geometric set, and let hh be a step function on S.S. Suppose P={S1,...,Sn}P={S1,...,Sn} and Q={T1,...,Tm}Q={T1,...,Tm} are two partitions of SS for which h(z)h(z) is the constant aiai on Si0Si0 and h(z)h(z) is the constant bjbj on Tj0.Tj0. Then

i = 1 n a i A ( S i ) = j = 1 m b j A ( T j ) . i = 1 n a i A ( S i ) = j = 1 m b j A ( T j ) .
(4)

Proof

We know by part (d) of (Reference) that the intersection of two geometric sets is itself a geometric set. Also, for each fixed index j,j, we know that the sets {TjSi0}{TjSi0} are pairwise disjoint. Then, by (Reference), we have that A(Tj)=i=1nA(TjSi).A(Tj)=i=1nA(TjSi). Similarly, for each fixed i,i, we have that A(Si=j=1mA(TjSi).A(Si=j=1mA(TjSi). Finally, for each pair ii and j,j, for which the set Tj0Si0Tj0Si0 is not empty, choose a point zi,jTj0Si0,zi,jTj0Si0, and note that ai=h(zi,j)=bj,ai=h(zi,j)=bj, because zi,jzi,j belongs to both Si0Si0 and Tj0.Tj0.

With these observations, we then have that

i = 1 n a i A ( S i ) = i = 1 n a i j = 1 m A ( T j S i ) = i = 1 n j = 1 m a i A ( T j S i ) = i = 1 n j = 1 m h ( z i , j ) A ( T j S i ) = i = 1 n j = 1 m b j A ( T j S i ) = j = 1 m i = 1 n b j A ( T j S i ) = j = 1 m b j i = 1 n A ( T j S i ) = j = 1 m b j A ( T j ) , i = 1 n a i A ( S i ) = i = 1 n a i j = 1 m A ( T j S i ) = i = 1 n j = 1 m a i A ( T j S i ) = i = 1 n j = 1 m h ( z i , j ) A ( T j S i ) = i = 1 n j = 1 m b j A ( T j S i ) = j = 1 m i = 1 n b j A ( T j S i ) = j = 1 m b j i = 1 n A ( T j S i ) = j = 1 m b j A ( T j ) ,
(5)

which completes the proof.

OK, the first consistency condition is satisfied. Moving right along:

Definition 3:

Let hh be a step function on a closed geometric set S.S. Define the integral of hh over the geometric set SS by the formula

S h = S H ( z ) d z = i = 1 n a i A ( S i ) , S h = S H ( z ) d z = i = 1 n a i A ( S i ) ,
(6)

where S1,...,SnS1,...,Sn is a partition of SS for which hh is the constant aiai on the interior Si0Si0 of the set Si.Si.

Just as in the case of integration on an interval, before checking the second consistency result, we need to establish the following properties of the integral of step functions.

Theorem 3

Let H(S)H(S) denote the vector space of all step functions on the closed geometric set S.S. Then the assignment hhhh of H(S)H(S) into RR has the following properties:

  1.  (Linearity) H(S)H(S) is a vector space, and S(h1+h2)=Sh1+Sh2,S(h1+h2)=Sh1+Sh2, and Sch=cShSch=cSh for all h1,h2,hH(S),h1,h2,hH(S), and for all real numbers c.c.
  2. If h=i=1nciχSih=i=1nciχSi is a linear combination of indicator functions of geometric sets that are subsets of S,S, then h=i=1nciA(Si).h=i=1nciA(Si).
  3.  (Positivity) If h(z)0h(z)0 for all zS,zS, then Sh0.Sh0.
  4.  (Order-preserving) If h1h1 and h2h2 are step functions on SS for which h1(z)h2(z)h1(z)h2(z) for all zS,zS, then Sh1Sh2.Sh1Sh2.

Proof

Suppose h1h1 is constant on the elements of a partition P={Si}P={Si} and h2h2 is constant on the elements of a partition Q={Tj}.Q={Tj}. Let VV be the partition of the geometric set SS whose elements are the sets {Uk}={Si0Tj0}.{Uk}={Si0Tj0}. Then both h1h1 and h2h2 are constant on the elements UkUk of V,V, so that h1+h2h1+h2 is also constant on these elements. Therefore, h1+h2h1+h2 is a step function, and

( h 1 + h 2 ) = k ( a k + b k ) A ( U k ) = k a k A ( U k ) + k b k A ( U k ) = h 1 + h 2 , ( h 1 + h 2 ) = k ( a k + b k ) A ( U k ) = k a k A ( U k ) + k b k A ( U k ) = h 1 + h 2 ,
(7)

and this proves the first assertion of part (1).

The proof of the other half of part (1), as well as parts (2), (3), and (4), are totally analogous to the proofs of the corresponding parts of (Reference), and we omit the arguments here.

Now for the other necessary consistency condition:

Theorem 4

let SS be a closed geometric set in the plane.

  1. If {hn}{hn} is a sequence of step functions that converges uniformly to a function ff on S,S, then the sequence {Shn}{Shn} is a convergent sequence of real numbers.
  2. If {hn}{hn} and {kn}{kn} are two sequences of step functions on SS that converge uniformly to the same function f,f, then
    limShn=limSkn.limShn=limSkn.
    (8)

Exercise 1

Prove Theorem 4. Mimic the proofs of (Reference) and (Reference).

Definition 4:

If ff is a real-valued function on a closed geometric set SS in the plane, then ff is integrable on S if it is the uniform limit of a sequence {hn}{hn} of step functions on S.S.

We define the integral of an integrable function ff on SS by

S f S f ( z ) d z = lim S h n , S f S f ( z ) d z = lim S h n ,
(9)

where {hn}{hn} is a sequence of step functions on SS that converges uniformly to f.f.

Theorem 5

Let SS be a closed geometric set in the plane, and let I(S)I(S) denote the set of integrable functions on S.S. Then:

  1.  I(S)I(S) is a vector space of functions.
  2. If ff and gI(S),gI(S), and one of them is bounded, then fgI(S).fgI(S).
  3. Every step function is in I(S).I(S).
  4. If ff is a continuous real-valued function on S,S, then ff is in I(S).I(S). That is, every continuous real-valued function on SS is integrable on S.S.

Exercise 2

  1. Prove Theorem 5. Note that this theorem is the analog of (Reference), but that some things are missing.
  2. Show that integrable functions on SS are not necessarily bounded; not even step functions have to be bounded.
  3. Show that, if fI(S),fI(S), and gg is a function on SS for which f(x,y)=g(x,y)f(x,y)=g(x,y) for all (x,y)(x,y) in the interior S0S0 of S,S, then gI(S).gI(S). That is, integrable functions on SS can do whatever they like on the boundary.

Theorem 6

Let SS be a closed geometric set. The assignment ffff on I(S)I(S) satisfies the following properties.

  1.  (Linearity) I(S)I(S) is a vector space, and S(αf+βg)=αSf+βSgS(αf+βg)=αSf+βSg for all f,gI(S)f,gI(S)and α,βR.α,βR.
  2.  (Positivity) If f(z)0f(z)0 for all zS,zS, then Sf0.Sf0.
  3.  (Order-preserving) If f,gI(S)f,gI(S) and f(z)g(z)f(z)g(z) for all zS,zS, then SfSg.SfSg.
  4. If fI(S),fI(S), then so is |f|,|f|, and |Sf|S|f|.|Sf|S|f|.
  5. If ff is the uniform limit of functions fn,fn, each of which is in I(S),I(S), then fI(S)fI(S) and Sf=limSfn.Sf=limSfn.
  6. Let {un}{un} be a sequence of functions in I(S),I(S), and suppose that for each nn there is a number mn,mn, for which |un(z)|mn|un(z)|mn for all zS,zS, and such that the infinite series mnmn converges. Then the infinite series unun converges uniformly to an integrable function, and Sun=Sun.Sun=Sun.
  7. If fI(S),fI(S), and {S1,...,Sn}{S1,...,Sn} is a partition of S,S, then fI(Si)fI(Si) for all i,i, and
    S=i=1nSif.S=i=1nSif.
    (10)

Exercise 3

Prove Theorem 6. It is mostly the analog to (Reference). To see the last part, let hihi be the step function that is identically 1 on Si;Si; check that hifI(Si);hifI(Si); then examine iSfhi.iSfhi.

Of course, we could now extend the notion of integrability over a geometric set SS to include complex-valued functions just as we did for integrability over an interval [a,b].[a,b]. However, real-valued functions on geometric sets will suffice for the purposes of this book.

We include here, to be used later in (Reference), a somewhat technical theorem about constructing partitions of a geometric set.

Theorem 7

Let S1,...,SnS1,...,Sn be closed, nonoverlapping, geometric sets, all contained in a geometric set S.S. Then there exists a partition S^1,...,S^MS^1,...,S^M of SS such that for 1in1in we have Si=S^i.Si=S^i. In other words, the sisi's are the first nn elements of a partition of S.S.

Proof

Suppose SS is determined by the interval [a,b][a,b] and the two bounding functions uu and l.l. We prove this theorem by induction on n.n.

If n=1,n=1, let S1S1 be determined by the interval [a1,b1][a1,b1] and the two bounding functions u1u1 and l1.l1. Set S^1=S1,S^1=S1, and define four more geometric sets S^2,...,S^5S^2,...,S^5 as follows:

  1.  S^2S^2 is determined by the interval [a,a1][a,a1] and the two bounding functions uu and ll restricted to that interval.
  2.  S3S3 is determined by the interval [a1,b1][a1,b1] and the two bounding functions uu and u1u1 restricted to that interval.
  3.  S4S4 is determined by the interval [a1,b1][a1,b1] and the two bounding functions ll and l1l1 restricted to that interval.
  4.  S^5S^5 is determined by the interval [b1,b][b1,b] and the two bounding functions uu and ll restricted to that interval.

Observe that the five sets S^1,S^2,...,S^5S^1,S^2,...,S^5 constitute a partition of the geometric set S,S, proving the theorem in the case n=1.n=1.

Suppose next that the theorem is true for any collection of nn sets satisfying the hypotheses. Then, given S1,...,Sn+1S1,...,Sn+1 as in the hypothesis of the theorem, apply the inductive hypothesis to the nn sets S1,...,SnS1,...,Sn to obtain a partition T1,...,TmT1,...,Tm of SS for which Ti=SiTi=Si for all 1in.1in. For each n+1im,n+1im, consider the geometric set Si'=Sn+1TiSi'=Sn+1Ti of the geometric set Ti.Ti. We may apply the case n=1n=1 of this theorem to this geometric set to conclude that Si'Si' is the first element Si,1'Si,1' of a partition {Si,1',Si,2',...,Si,mi'}{Si,1',Si,2',...,Si,mi'} of the geometric set Ti.Ti.

Define a partition {S^k}{S^k} of SS as follows: For 1kn,1kn, set S^k=Tk.S^k=Tk. Set S^n+1=i=n+1mSi,1'=Sn+1.S^n+1=i=n+1mSi,1'=Sn+1. And define the rest of the partition {S^k}{S^k} to be made up of the remaining sets Si,j'Si,j' for n+1imn+1im and 2jmi.2jmi. It follows directly that this partition {S^k}{S^k} satisfies the requirements of the theorem.

Exercise 4

Let S1,...,SnS1,...,Sn be as in the preceding theorem. Suppose SkSk is determined by the interval [ak,bk][ak,bk] and the two bounding functions ukuk and lk.lk. We will say that SkSk is “below” Sj,Sj, equivalently SjSj is “above” Sk,Sk, if there exists a point xx such that uk(x)<lj(x).uk(x)<lj(x). Note that this implies that x[ak,bk][aj,bj].x[ak,bk][aj,bj].

  1. Suppose SkSk is below Sj,Sj, and suppose (z,yk)Sk(z,yk)Sk and (z,yj)Sj.(z,yj)Sj. Show that yj>yk.yj>yk. That is, if SkSk is below Sj,Sj, then no part of SkSk can be above Sj.Sj.
  2. Suppose S2S2 is below S1S1 and S3S3 is below S2.S2. Show that no part of S3S3 can be above S1.S1. HINT: By way of contradiction, let x1[a1,b1]x1[a1,b1] be such that u2(x1)<l1(x1);u2(x1)<l1(x1); let x2[a2,b2]x2[a2,b2] be such that u3(x2)<l2(x2);u3(x2)<l2(x2); and suppose x3[a3,b3]x3[a3,b3] is such that u1(x3)<l3(x3).u1(x3)<l3(x3). Derive contradictions for all possible arrangements of the three points x1,x2,x1,x2, and x3.x3.
  3. Prove that there exists an index k0k0 such that Sk0Sk0 is minimal in the sense that there is no other SjSj that is below Sk0.Sk0. HINT: Argue by induction on n.n. Thus, let {Tl}{Tl} be the collection of all SkSk's that are below S1,S1, and note that there are at most n-1n-1 elements of {Tl}.{Tl}. By induction, there is one of the TlTl's, i.e., an Sk0Sk0 that is minimal for that collection. Now, using part (b), show that this Sk0Sk0 must be minimal for the original collection.

There is one more concept about integrating over geometric sets that we will need in later chapters. We have only considered sets that are bounded on the left and right by straight vertical lines and along the top and bottom by graphs of continuous functions y=u(x)y=u(x) and y=l(x).y=l(x). We equally well could have discussed sets that are bounded above and below by straight horizontal lines and bounded on the left and right by graphs of continuous functions x=l(y)x=l(y) and x=r(y).x=r(y). These additional sets do not provide anything particularly important, so we do not discuss them. However, there are times when it is helpful to work with geometric sets with the roles of horizontal and vertical reversed. We accomplish this with the following definition.

Definition 5:

Let SS be a subset of R2.R2. By the symmetric image of SS we mean the set S^S^ of all points (x,y)R2(x,y)R2 for which the point (y,x)S.(y,x)S.

The symmetric image of a set is just the reflection of the set across the y=xy=x line in the plane. Note that the symmetric image of the rectangle [a,b]×[c,d][a,b]×[c,d] is again a rectangle, [c,d]×[a,b],[c,d]×[a,b], and therefore the area of a rectangle is equal to the area of its symmetric image. This has the implication that if the symmetric image of a geometric set is also a geometric set, then they both have the same area. The symmetric image of a geometric set doesn't have to be a geometric set itself. For instance, consider the examples suggested in part (b) of (Reference). But clearly rectangles, triangles, and circles have this property, for their symmetric images are again rectangles, triangles, and circles. For a geometric set, whose symmetric image is again a geometric set, there are some additional computational properties of the area of SS as well as the integral of functions over S,S, and we present them in the following exercises.

Exercise 5

Suppose SS is a closed geometric set, which is determined by a closed interval [a,b][a,b] and two bounding functions u(x)u(x) and l(x).l(x). Suppose the symmetric image S^S^ of SS is also a closed geometric set, determined by an interval [a^,b^][a^,b^] and two bounding functions u^(x)u^(x) and l^(x).l^(x).

  1. Make up an example to show that the numbers a^a^ and b^b^ need not have anything to do with the numbers aa and b,b, and that the functions u^u^ and l^l^ need not have anything to do with the functions uu and l.l.
  2. Prove that SS and S^S^ have the same area. HINT: use the fact that the area of a geometric set is approximately equal to the sum of the areas of certain rectangles, and then use the fact that the area of the symmetric image of a rectangle is the same as the area of the rectangle.
  3. Show that for every point (x,y)S,(x,y)S, we must have a^yb^,a^yb^, and for every such y,y, we must have l^(y)xu^(y).l^(y)xu^(y). HINT: If (x,y)S,(x,y)S, then (y,x)S^.(y,x)S^.
  4. (d) Prove that the area A(S)A(S) of SS is given by the formula
    A(S)=abl(x)u(x)1dydx=a^b^l^(y)u^(y)1dxdy.A(S)=abl(x)u(x)1dydx=a^b^l^(y)u^(y)1dxdy.
    (11)
    (See (Reference).)
  5. Let SS be the right triangle having vertices (a,c),(b,c),(a,c),(b,c), and (b,d),(b,d), where d>c.d>c. Describe the symmetric image of S;S; i.e., find the corresponding a^,b^,u^,a^,b^,u^, and l^.l^. Use part (d) to obtain the following formulas for the area of S:S:
    A(S)=abcd+t-ba-b(c-d)1dsdt=cdb+t-dc-d(a-b)1dsdt.A(S)=abcd+t-ba-b(c-d)1dsdt=cdb+t-dc-d(a-b)1dsdt.
    (12)

Exercise 6

  1. Prove that if S1S1 and S2S2 are geometric sets whose symmetric images are again geometric sets, then the symmetric image of the geometric set S1S2S1S2 is also a geometric set.
  2. Suppose TT is a closed geometric set that is contained in a closed geometric set S.S. Assume that both the symmetric images T^T^ and S^S^ are also geometric sets. If S^S^ is determined by an interval [a^,b^][a^,b^] and two bounding functions u^u^ and l^,l^, prove that
    A(T)=a^b^l^(s)u^(s)χT(t,s)dtds,A(T)=a^b^l^(s)u^(s)χT(t,s)dtds,
    (13)
    where χTχT is the indicator function of the set T;T; i.e., χT(t,s)=1χT(t,s)=1 if (t,s)T,(t,s)T, and χT(t,s)=0χT(t,s)=0 if (t,s)T.(t,s)T. HINT: See the proof of (Reference), give names to all the intervals and bounding functions, and in the end use part (d) of the preceding exercise.
  3. Suppose {Si}{Si} is a partition of a geometric set S,S, and suppose the symmetric images of SS and all the SiSi's are also geometric sets. Suppose hh is a step function that is the constant aiai on the element Si0Si0 of the partition {Si}.{Si}. Prove that Sh=i=1naiSχSi0,Sh=i=1naiSχSi0, and therefore that
    Sh=abl(t)u(t)h(t,s)dsdt=a^b^l^(s)u^(s)h(t,s)dtds.Sh=abl(t)u(t)h(t,s)dsdt=a^b^l^(s)u^(s)h(t,s)dtds.
    (14)
    HINT: Use part (b).
  4. Let SS be a geometric set whose symmetric image S^S^ is also a geometric set, and suppose ff is a continuous function on S.S. Show that
    Sf=abl(t)u(t)f(t,s)dsdt=a^b^l^(s)u^(s)f(t,s)dtds.Sf=abl(t)u(t)f(t,s)dsdt=a^b^l^(s)u^(s)f(t,s)dtds.
    (15)
    HINT: Make use of the fact that the step functions constructed in Theorem 1 satisfy the assumptions of part (c). Then take limits.
  5. Let SS be the triangle in part (e) of the preceding exercise. If ff is a continuous function on S,S, show that the integral of ff over SS is given by the formulas
    Sf=abcd+t-ba-b(c-d)f(t,s)dsdt=cdb+s-dc-d(a-b)f(t,s)dtds.Sf=abcd+t-ba-b(c-d)f(t,s)dsdt=cdb+s-dc-d(a-b)f(t,s)dtds.
    (16)

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'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks