I have a problem with improper integral in my book. You need to know the following information.
1) I am using this definition of improper/extended integrals over open sets:
Let $A$ be an open set in $R^n$; let $f : A \rightarrow R$ be a continuous
function. If $f$ is non-negative on $A$, we define the $(extended) \ integral \ of \ f \ over \ A$, denoted $\int_A f$, to be the supremum of the numbers $\int_D f$, as $D$ ranges over all compact rectifiable subsets of $A$, provided this supremum exists. In this case, we say that $f \ is \ integrable \ over \ A$ (in the extended sense).
2) I am also using this $change \ of \ variables$ version:
Let $g: A \rightarrow B$ be a diffeomorphism of open sets in $R^n$. Let $f : B \rightarrow R$ be a continuous function. Then $f$ is integrable over $B$ if and only if the function
$(fog)|detDg|$ is integrable over $A$; in this case,
$$\int_B f=\int_A (fog)|detDg|$$
I am trying to understand this example from my book:
Suppose we wish to integrate the function $x^2y^2$ over the open set $W$={$(x,y)$ | $x^2+y^2<a^2$}. Here the use of polar coordinates is a bit more tricky. The polar coordinate transformation $g$ does not in this case define a diffeomorphism of an open set in the $(r,θ)$ plane with $W$. However, $g$ does define a diffeomorphism of the open set $U = (0, α)$ x $(0, 2π)$ with the open set $V$ = {$(x, y)$ | $x^2 + y^2 < α^2$ and $x < 0$ if $y = 0$} of $R^2$.
The set $V$ consists of $W$ with the non-negative $x$-axis
deleted. Because the non-negative $x$-axis has $measure \ 0$ (and even $content \ 0$) , $\int_W x^2y^2$ exists and
$\int_W x^2y^2=\int_V x^2y^2$.
The latter can be expressed as an integral over $U$, by use of the polar coordinate transformation.
MY QUESTION IS why is this claim true?: Let $f:W \rightarrow R$ be continuous (and possibly unbounded) and $W$ is an open subset of $R^n$ such that it is the union of the open set $V$ and the set with measure $0$ $E$. If the extended integral $\int_V f$ exists, then the extended integral $\int_W f$ also exists and
$$\int_W f=\int_V f$$.
Is it possible to prove this at least in the case that $E$ has $content \ 0$?
Thanks in advance for any help!
Best Answer
Theorem 15.6 in Analysis on Manifolds by Munkres states
I will add that each $O_j$ can be chosen as the interior of a compact rectifiable set $\bar{O}_j$ and, since, $f$ is continuous and bounded on $\bar{O}_j$ the integrals $\int_{\bar{O}_j} f = \int_{O_j} f$ exist properly. (The existence of $\int_{O_j} f$ over the interior and equality to $\int_{\bar{O}_j} f$ follows from Theorem 13.6 in Munkres.)
I think I have found a way to prove this statement for the case that $E$ has $content \ 0$. For this problem we are given that $W = V \cup E$ where $W$ and $V$ are open and $E$ is of measure $0$. We can assume that $V \cap E = \emptyset$ (since this seems to be the right way to state this theorem), so that $V=W-E=W \cap E^c$. We are also given that $f$ is continuous on $W$ and the extended integral $\int_V f$ exists.
Take a sequence $O_1 \subset O_2 \subset \ldots$ of open sets such that $W = \cup_{j=1}^\infty O_j$. Since $O_j \subset W$, then $O_j=O_j \cap W$. Thus the sets $\hat{O}_j = O_j \cap E^c=(O_j \cap W) \cap E^c=O_j \cap (W \cap E^c)=O_j \cap V$ are also open and and nested with union $V = \cup_{j=1}^\infty \hat{O}_j$.
We know that if $A$ has content $0$ then it is $rectifiable$ (it is bounded and its boundary has measure $0$) and it has measure $0$. Thus it can be proven (by Theorem 11.3) that $\int_A f$ exists and equals $0$.
Now since $O_j \cap E \subset E$ and $E$ has content $0$, then $O_j \cap E$ has content $0$. It follows that $\int_{O_j \cap E} |f|=0$. Then by Corollary 13.4
$$\int_{O_j} |f| = \int_{\hat{O_j}} |f| + \int_{O_j \cap E} |f| = \int_{\hat{O_j}} |f|$$.
Thus the sequence $\int_{O_j} |f|$ is bounded and the extended integral $\int_W f$ exists. Furthermore, as shown for $|f|$,
$$\int_{O_j} f = \int_{\hat{O_j}} f,$$
and, it follows that,
$$\int_W f = \lim_{j \to \infty}\int_{O_j} f = \lim_{j \to \infty}\int_{\hat{O_j}} f = \int_V f$$