Let $S_1,S_2$ be bounded sets in $\mathbb{R}^n$; let $f:S_1\cup S_2\rightarrow\mathbb{R}$ 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 know there is the result that $f$ is integrable over $S_1\cup S_2$ and over $S_1\cap S_2$, and also the formula $$\int_{S_1\cup S_2}f=\int_{S_1}f+\int_{S_2}f-\int_{S_1\cap S_2}f.$$
But I can't see how to apply this result here.
Also, I tried looking at the proof of this result. For nonnegative $f$ it uses $f_{S_1\cup S_2}(x)=\max\{f_{S_1}(x),f_{S_2}(x)\}$ (where $f_A(x)=f(x)$ if $x\in A$ and $=0$ if $x\not\in A$.), and then the fact that the max function of two integrable functions is integrable. Then for general $f$ it uses $f_+(x)=\max\{f(x),0\}, f_-(x)=\max\{-f(x),0\}$, and $f(x)=f_+(x)-f_-(x)$, and apply linearity.
Might any of this be helpful toward the problem?
Best Answer
Note that $$S_1=(S_1\smallsetminus S_2)\sqcup(S_1\cap S_2)$$