- #1

Bill2500

- 10

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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!

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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