Integral of Integrable Function over Null Set

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


Proof


Sources