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First, observe that: \begin{align} A & = & (A\cap B^{c})\cup (A\cap B) \\ & = & ((A\cup B)\cap B^c)\cup(A\cap B) \\ & = & (A\setminus B)\cup (A\cap B) \end{align}

Second observe that: \begin{align} B & = & (B\cap A^{c})\cup (B\cap A) \\ & = & ((B\cup A)\cap A^c)\cup(B\cap A) \\ & = & (B\setminus A)\cup (B\cap A) \end{align}

Using the Caratheodory condition since both $A$ and $B$ are measurable sets: \begin{align} m(A\cup B) & = & m((A\cup B)\cap A)+m((A\cup B)\cap A^c) \\ & = & m(A) + m(B \setminus A) \end{align}

Again applying Caratheodory condition:

There is something missing here. How do I justify this? What property of measure gives me that $m(A\cap B)= m(B)-m(B\setminus A)$? \begin{align} m(A\cap B) & = & m((A\cap B)\cap A)+m((A\cup B)\cap A^c) \\ & = & m((A\cap B)\cap A) \\ & = & m(B)-m(B \setminus A) \end{align}

Therefore: $$m(A\cup B) +m(A\cap B) = m(A) + m(B)$$

Thank you very much.

PS this is from Carothers chapter 16 on measurable sets.

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    For your question : $B\backslash A$ and $A\cap B$ are disjoint ! The result follow. – 2017-02-11
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    Oh yes of course they are disjoint. That's it. – 2017-02-11

2 Answers 2

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You seem to have it. You may as well assume $m(B) < \infty$ since otherwise the equality holds trivially.

Since $A$ is measurable you have $m(B) = m(B \cap A) + m(B \cap A^c)$. This gives you $m(A \cap B) = m(B) - m(B \cap A^c) = m(B) - m(B \setminus A)$.

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Another proof: clearly $1_{A \cup B} + 1_{A \cap B} = 1_A + 1_B$, where $1_X(x)=1$ if $x \in X$ and $0$ otherwise. Now integrate.

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    haha this is cool... I did not think of it this way. But I think I see what you are saying. – 2017-02-11