How to show that $m(E_1 \cap E_2)+m(E_1 \cup E_2)=m(E_1)+m(E_2)$?

Question

Let $E_1$ and $E_2$ measurable sets.

Show that $m(E_1 \cap E_2)+m(E_1 \cup E_2)=m(E_1)+m(E_2)$

if $E_1$ is measurable, for any set A, $m^*(A)=m^*(A\cap E_1)+m^*(A \cap E_1^c)$

how can I start? (sorry my jvs isnt working I cant add correct tag)

Answer 1

(Assumption: $m$ denotes measure, $m^*$ denotes outer measure.)

You can decompose $$ E_1\cup E_2=(E_1\backslash E_2)\cup(E_2\backslash E_1)\cup(E_1\cap E_2)\tag{*} $$ then use the additive property (3 times).

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To be more explicit: \begin{gather*} E_1=(E_1\backslash E_2)\cup(E_1\cap E_2)\implies m(E_1)=m(E_1\backslash E_2)+m(E_1\cap E_2),\\ E_2=(E_2\backslash E_1)\cup(E_1\cap E_2)\implies m(E_2)=m(E_2\backslash E_1)+m(E_1\cap E_2). \end{gather*} So $$ m(E_1)+m(E_2)=m(E_1\cap E_2)+[m(E_1\backslash E_2)+m(E_2\backslash E_1)+m(E_1\cap E_2)]. $$ Now use the decomposition given in (*).

Answer 2

$$E_{1}\cup E_{2}=\left(E_{1}\backslash E_{2}\right)\cup\left(E_{2}\backslash E_{1}\right)\cup\left(E_{1}\cap E_{2}\right)$$ Consequently: $$m\left(E_{1}\cup E_{2}\right)+m\left(E_{1}\cap E_{2}\right)=m\left(E_{1}\backslash E_{2}\right)+m\left(E_{2}\backslash E_{1}\right)+2m\left(E_{1}\cap E_{2}\right)$$

$$E_{1}=\left(E_{1}\backslash E_{2}\right)\cup\left(E_{1}\cap E_{2}\right)\text{ and } E_{2}=\left(E_{2}\backslash E_{1}\right)\cup\left(E_{1}\cap E_{2}\right)$$ Consequently: $$m\left(E_{1}\right)+m\left(E_{2}\right)=m\left(E_{1}\backslash E_{2}\right)+m\left(E_{2}\backslash E_{1}\right)+2m\left(E_{1}\cap E_{2}\right)$$

So the RHS of the equations is the same. This allows the conclusion: $$m\left(E_{1}\cup E_{2}\right)+m\left(E_{1}\cap E_{2}\right)=m\left(E_{1}\right)+m\left(E_{2}\right)$$

Answer 3

Integrate with respect to the measure $m$ the pointwise identity $$\mathbf 1_{E_1\cup E_2}+\mathbf 1_{E_1\cap E_2}=\mathbf 1_{E_1}+\mathbf 1_{E_2}.$$