Integral of Integrable Function over Null Set

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

Let $f: X \to \overline \R$ be a $\mu$-integrable function.

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

Then:


 * $\ds \int_N f \rd \mu = 0$

where $\ds \int_N$ signifies an integral over $N$.