Munkres-Analysis on Manifolds: Extended Integrals

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Main Question or Discussion Point

I am studying Analysis on Manifolds by Munkres. He introduces improper/extended integrals over open set the following way: Let A be an open set in R^n; let f : A -> R be a continuous function. If f is non-negative on A, we define the (extended) integral of f over A, as the supremum of all the numbers "integral of f over D" , as D ranges over all compact rectifiable (boundary of D has measure 0) subsets of A, provided this supremum exists. In this case, we say that f is integrable over A (in the extended sense).

In section 17 example 4 he uses the change of variables theorem for extended integrals over open sets to calculate the integral of a continuous non-negative function f over an open circle of radius a (lets call it W). To do so, he uses polar coordinates and integrates the "polar coordinate" function g over U=(0,a)x(0,2π).

But g(U)=V={(x,y) in W| x<0 if y=0} which is a subset of W, different from W. So "the integral of f over V" exists. Finally because W-V has measure 0, he claims that "the integral of f over W" also exists and "the integral of f over W"="the integral of f over V" (these are extended integrals over open set).

I don't understand how I can prove the claim: If f is continuous and if W is open and it is the union of the open set V and a set E with measure 0 and the extended " integral of f over V" exists, then "the integral of f over W" exists and they are equal (especially since the function f may not be bounded over W).

Thank you in advance for any help!
 
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  • #2
andrewkirk
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I don't understand how I can prove the claim: If f is continuous and if W is open and it is the union of the open set V and a set E with measure 0 and the extended " integral of f over V" exists, then "the integral of f over W" exists and they are equal (especially since the function f may not be bounded over W).
The integral of any function over a set of measure zero is zero, see proof here. Hence the integral over any rectifiable subset of a set of measure zero must also be zero.

Since integrals on disjoint sets are additive, the complements and intersections of finite numbers of rectifiable sets are rectifiable, ##W=V\cup E##, V and E are disjoint, and V and E are both rectifiable, we can write an integral over any rectifiable subset of W as the integral over ##W\cap V## plus the integral over ##W\cap E##. The latter is always zero. So the supremum of integrals of rectifiable subsets of W cannot be greater than the supremum of integrals of rectifiable subsets of V.

Since any rectifiable subset of V is also a rectifiable subset of W, the opposite inequality also applies. So the two suprema are equal.

PS Are you sure that posting that link to an online copy of a textbook doesn't breach copyright rules? Unless you are very sure, I suggest removing the link, or requesting a moderator to do it if the post is no longer editable. It is OK to post images of a page or two, but generally posting a whole book will violate copyright.
 
  • #3
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The integral of any function over a set of measure zero is zero, see proof here. Hence the integral over any rectifiable subset of a set of measure zero must also be zero.

Since integrals on disjoint sets are additive, the complements and intersections of finite numbers of rectifiable sets are rectifiable, ##W=V\cup E##, V and E are disjoint, and V and E are both rectifiable, we can write an integral over any rectifiable subset of W as the integral over ##W\cap V## plus the integral over ##W\cap E##. The latter is always zero. So the supremum of integrals of rectifiable subsets of W cannot be greater than the supremum of integrals of rectifiable subsets of V.

Since any rectifiable subset of V is also a rectifiable subset of W, the opposite inequality also applies. So the two suprema are equal.

PS Are you sure that posting that link to an online copy of a textbook doesn't breach copyright rules? Unless you are very sure, I suggest removing the link, or requesting a moderator to do it if the post is no longer editable. It is OK to post images of a page or two, but generally posting a whole book will violate copyright.
I don't know how I can upload a picture so I found a link of the book and included it.
 
  • #4
andrewkirk
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Yes, it's not obvious, and I often forget how to do it. Fortunately, I had to do one just the other day, so I remember for now.

Two buttons to the right of the 'Post Reply' button is a button marked Upload. Click on that and it allows you to upload an image from your computer. It will then appear in the post. If you choose Thumbnail it will look small, but expand to full size when clicked. Otherwise it will always be fully expanded in your post. I feel the former is easier to read when referencing a text.
 
  • #5
10
2
I am studying Analysis on Manifolds by Munkres. He introduces improper/extended integrals over open set the following way: Let A be an open set in R^n; let f : A -> R be a continuous function. If f is non-negative on A, we define the (extended) integral of f over A, as the supremum of all the numbers "integral of f over D" , as D ranges over all compact rectifiable (boundary of D has measure 0) subsets of A, provided this supremum exists. In this case, we say that f is integrable over A (in the extended sense).

In section 17 example 4 he uses the change of variables theorem for extended integrals over open sets to calculate the integral of a continuous non-negative function f over an open circle of radius a (lets call it W). To do so, he uses polar coordinates and integrates the "polar coordinate" function g over U=(0,a)x(0,2π).

But g(U)=V={(x,y) in W| x<0 if y=0} which is a subset of W, different from W. So "the integral of f over V" exists. Finally because W-V has measure 0, he claims that "the integral of f over W" also exists and "the integral of f over W"="the integral of f over V" (these are extended integrals over open set).

I don't understand how I can prove the claim: If f is continuous and if W is open and it is the union of the open set V and a set E with measure 0 and the extended " integral of f over V" exists, then "the integral of f over W" exists and they are equal (especially since the function f may not be bounded over W).

Thank you in advance for any help!
extended.jpg
extended.jpg
 

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  • #6
10
2
I am studying Analysis on Manifolds by Munkres. He introduces improper/extended integrals over open set the following way: Let A be an open set in R^n; let f : A -> R be a continuous function. If f is non-negative on A, we define the (extended) integral of f over A, as the supremum of all the numbers "integral of f over D" , as D ranges over all compact rectifiable (boundary of D has measure 0) subsets of A, provided this supremum exists. In this case, we say that f is integrable over A (in the extended sense).

In section 17 example 4 he uses the change of variables theorem for extended integrals over open sets to calculate the integral of a continuous non-negative function f over an open circle of radius a (lets call it W). To do so, he uses polar coordinates and integrates the "polar coordinate" function g over U=(0,a)x(0,2π).

But g(U)=V={(x,y) in W| x<0 if y=0} which is a subset of W, different from W. So "the integral of f over V" exists. Finally because W-V has measure 0, he claims that "the integral of f over W" also exists and "the integral of f over W"="the integral of f over V" (these are extended integrals over open set).

I don't understand how I can prove the claim: If f is continuous and if W is open and it is the union of the open set V and a set E with measure 0 and the extended " integral of f over V" exists, then "the integral of f over W" exists and they are equal (especially since the function f may not be bounded over W).

Thank you in advance for any help!
Yes, it's not obvious, and I often forget how to do it. Fortunately, I had to do one just the other day, so I remember for now.

Two buttons to the right of the 'Post Reply' button is a button marked Upload. Click on that and it allows you to upload an image from your computer. It will then appear in the post. If you choose Thumbnail it will look small, but expand to full size when clicked. Otherwise it will always be fully expanded in your post. I feel the former is easier to read when referencing a text.
It does not add it to my post. Why?
 

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