Show that measure of symmetric difference $= \int_{E}\vert \chi_{A} - \chi_{B}\vert$

Question

I have just proven that the Nikodym (pseudo)metric $\rho(A,B)$ is a pseudometric (i.e., satisfies all the metric axioms, except that $\rho(A,B)$ can be $0$ even if $A \neq B$), and now I need to show that $\mathbf{\rho(A,B)=\int_{E}\vert\chi_{A}-\chi_{B}\vert}$.

A little background:

Let $E$ be a Lebesgue measurable set of real numbers of finite measure. Let $X$ be the set of Lebesgue measurable subsets of $E$, and let $m$ denote the Lebesgue measure. For $A$, $B$ $\in X$, define $\rho(A,B) = m(A \Delta B) = m((A\backslash B) \cup (B \backslash A))$ (i.e., the symmetric difference of $A$ and $B$).

To show that $\rho(A,B)=\int_{E}\vert\chi_{A}-\chi_{B}\vert$, first off I assume we can apply countable additivity of the Lebesgue measure, since $A\backslash B$ and $B\backslash A$ are disjoint, to give us $\rho(A,B)=m(A \Delta B) = m[(A\backslash B) \cup (B \backslash A)] = m(A \backslash B) + m(B \backslash A)$, but I don't know where to go from there.

Please help!

Answer 1

$\left|\chi_{A}-\chi_{B}\right|=\chi_{A\triangle B}$ so that: $$\int\left|\chi_{A}-\chi_{B}\right|dm=\int\chi_{A\triangle B}dm=m\left(A\triangle B\right)=\rho\left(A,B\right)$$

edit:

Note that $\left|\chi_{A}-\chi_{B}\right|$ can only take values in $\{0,1\}$.

• If $x\in A\cap B$ then $\chi_{A}(x)=1\wedge\chi_{B}(x)=1$ so $\left|\chi_{A}-\chi_{B}\right|=0$

• If $x\in A\setminus B$ then $\chi_{A}(x)=1\wedge\chi_{B}(x)=0$ so $\left|\chi_{A}-\chi_{B}\right|=1$

• If $x\in B\setminus A$ then $\chi_{A}(x)=0\wedge\chi_{B}(x)=1$ so $\left|\chi_{A}-\chi_{B}\right|=1$

• If $x\notin A\cup B$ then $\chi_{A}(x)=0\wedge\chi_{B}(x)=0$ so $\left|\chi_{A}-\chi_{B}\right|=0$