Characteristic Function of Null Set is A.E. Equal to Zero

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $N$ be a $\mu$-null set.

Then:


 * $\chi_N = 0$ $\mu$-almost everywhere.

where $\chi_N$ is the characteristic function of $N$.

Proof
Let $x \in X$ be such that:


 * $\map {\chi_N} x \ne 0$

Then. since $\map {\chi_N} x \in \set {0, 1}$:


 * $\map {\chi_N} x = 1$

which is equivalent to:


 * $x \in N$

from the definition of a characteristic function.

So:


 * if $x \in X$ is such that $\map {\chi_N} x \ne 0$, then $x \in N$.

Since $N$ is a $\mu$-null set, we have:


 * $\chi_N = 0$ $\mu$-almost everywhere.