Let $S_1$ and $S_2$ be bounded sets in $\mathbb{R}^n$; let $f:S\to\mathbb{R}$ be a bounded function. Show that if $f$ is integrable over $S_1$ and $S_2$, then $f$ is integrable over $S_1-S_2$, and $\int_{S_1-S_2}f=\int_{S_1}f-\int_{S_1\cap S_2}f$.
I got this question from the link.
@positrón0802 said the in the link:
Recall that given a bounded set $A\subset \mathbb{R}^n$ and $g:A\to \mathbb{R},$ we define
$g_A:\mathbb{R}^n\to \mathbb{R}$ by
$$ g_A(x)=\begin{cases}g(x), &x\in A,\\0 & \text{otherwise.}\end{cases}$$
First consider the case in which $f$ is non-negative.
Now let $S=S_1\cup S_2$ and $T=S_1\cap S_2.$ If $f$ is integrable over $S_1$ and $S_2$ then, applying Lemma 13.2 (Munkres Analysis on Manifolds), it is integrable over $S$ and $T$ because $f_S(x)=\max\{f_{S_1}(x),f_{S_2}(x)\}$ and $f_T(x)=\min\{f_{S_1}(x),f_{S_2}(x)\}.$
Now show the following:
- 1) Show that $f_{S_1-S_2}(x)=\max\{0,f_{S_1}(x)-f_{S_2}(x)\}.$ For this you can consider the cases $x\in T,$ $x\in S_2,$ $x\in S_1-S_2,$ $x\not\in S.$ From this follows that $f$ is integrable over $S_1-S_2.$
- 2) Show that $f_{S_1-S_2}(x)=f_{S_1}(x)-f_T(x).$ Again you can consider the cases $x\in T,$ $x\in S_2,$ $x\in S_1-S_2,$ $x\not\in S.$ This complete the result if $f$ is non-negative.
Finally, in the general case ($f$ not necessarily non-negative), set
$$ f_+(x)=\max\{f(x),0\} \hspace{.3cm}\text{ and } \hspace{.3cm}f_-(x)=\max\{-f(x),0\}.$$
Both are non-negative and $f(x)=f_+(x)-f_-(x).$
May you say that how can we show $1)$ and $2)$? Thanks…
Best Answer
For both 1) & 2) we can look at three cases that cover all relevant possibilities:
(i) $x\notin S_1 $: $f_{S_1}(x) = 0$, $f_{S_2}(x) = 0$, $f_{S_1 \setminus S_2} (x) = 0$.
(ii) $x \in S_1 \cap S_2$: $f_{S_1}(x) = f(x)$, $f_{S_2}(x) = f(x)$, $f_{S_1 \setminus S_2} (x) = 0$.
(iii) $x \in S_1 \setminus S_2$: $f_{S_1}(x) = f(x)$, $f_{S_2}(x) = 0$, $f_{S_1 \setminus S_2} (x) = f(x)$.
It is straightforward to check that the left and right hand side are equal is all cases.