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Suppose I have a function $f \in L^1(\mathbb{R})$.

I know that a few key properties hold on general Lebesgue functions, including additivity: for disjoint subsets $E_1, E_2$ of $\mathbb{R}$ we have that

$\int_{E_1 \cup E_2} f = \int_{E_1} f + \int_{E_2} f$

Suppose instead we have that, WLOG, $E_1 \subseteq E_2$.

Can it be said that the following is true:

$\int_{E_1 - E_2} f = \int_{E_1} f - \int_{E_2} f$

Is this trivial? What would be the best approach of proving this statement?

1 Answers 1

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First you should assume that $E_1,E_2$ are Lebesgue measurable subsets of $\mathbb{R}$ and $E_1\supset E_2$. With this, we have $$E_1-E_2=E_1\cap E_2^c$$ which is a measurable set, and $E_1$ can be written as a union of disjoint measurable sets: $$E_1=E_2\cup(E_1-E_2).$$ It follows that $$\int_{E_1}f=\int_{E_2}f+\int_{E_1-E_2}f.$$ This shows the claim.

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    Perhaps I'm missing something, but this yields $\int_{E_2}f=0$ if $E_1=\{\}$, no? I think $E_1=E_2\cup(E_1\setminus E_2)$ is wrong. – 2017-02-16
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    @YoTengoUnLCD Here we assume that $E_1\supseteq E_2$. I add that to my answer. – 2017-02-16