Integral of Integrable Function over Null Set

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

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

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

Then:


 * $\displaystyle \int_N f \, \mathrm d \mu = 0$

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