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 $N$.

Corollary

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

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

Then:

$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.

where $\chi_{X \setminus N}$ is the characteristic function of $X \setminus N$.

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.

$\blacksquare$