# How to show that $m(A \cup B)+m(A \cap B) = m(A)+m(B)$?

How to show that $m(A \cup B)+m(A \cap B) = m(A)+m(B)$ when $A \subset \mathbb{R}^n$ and $B \subset \mathbb{R}^n$ are measurable( in lebesgue sense). just hint.

-
which definition of lebesgue measure are you using? – Norbert Jan 28 '13 at 9:39
$A\cup B=A\cup (B\backslash A)$. – Michael Greinecker Jan 28 '13 at 9:44
– alvoutila Jan 28 '13 at 12:10
By the way: One insignificant guestion: If you would show that two sets are the same, then you should show that $A \subset B$ and $B \subset A$. Am I right? – alvoutila Jan 28 '13 at 22:47

Note that the union of two measurable sets is again measurable. Recall if $E$ is measurable then $m(S) = m(S\cap E) + m(S\cap E^c)$ for every set $S \subseteq \mathbb{R}$. Using this, we want to relate the two terms on the left of the desired equation to the two terms on the right. First note that

$$m(A\cup B) = m((A\cup B)\cap A) + m((A\cup B)\cap A^c) = m(A) + m(B\setminus A).$$

Now try to conceive how you could use the condition $m(S) = m(S\cap E) + m(S\cap E^c)$ to relate $m(A\cap B)$ and $m(B)$.

Added Later: Using the same ideas, you could also relate $m(A\cup B)$ to $m(B)$, and $m(A\cap B)$ to $m(A)$. Either way, you obtain the desired equation.

-
I thought that $m(S)=m(S \cap E) + m(S \cap E^c)$ means that E is measurable. Thus in your notation S would be test set( arbitrary). Am i right? – alvoutila Jan 28 '13 at 12:12
You are correct. I should have said $E$ is measurable and that the equation held for arbitrary $S$. – Michael Albanese Jan 28 '13 at 12:15
Shouldn't $(A \cup B)\cap A^c$ be same as $(A \cup B)\setminus A$ not $B\setminus A$? – alvoutila Jan 28 '13 at 15:47
Intersection distributes over union so $$(A\cup B)\cap A^c = (A\cap A^c)\cup(B\cap A^c) = \emptyset\cup(B\setminus A) = B\setminus A.$$ – Michael Albanese Jan 28 '13 at 16:03
Just curious: How you invented that $A \cup B = [(A \cup B) \cap A]\cup [(A \cup B)\cap A^c]$? – alvoutila Jan 28 '13 at 16:36

Note that $A\cup B$ is the disjoint union $(A\setminus B)\cup B$ or $(B\setminus A)\cup A$.

Besides $A\setminus B=A\setminus (A\cap B)$ and $A\cap B\subseteq A$, and similarly $B\setminus A$.

For the properties of the measure all these sets are measurable (given $A$ and $B$ are measurable).

-