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 $N$.

Proof

We have, from definition:

$\ds \int_N f \rd \mu = \int f \cdot \chi_N \rd \mu$

Note that if $x \in X \setminus N$, we have:

$\map {\chi_N} x = 0$

So, if:

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

we must have $x \in N$.

Since $N$ is a null set, this gives:

$f \cdot \chi_N = 0$ $\mu$-almost everywhere.

From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we therefore have:

$\ds \int \size {f \cdot \chi_N} \rd \mu = 0$

Then, from Triangle Inequality for Integrals: Corollary, we have:

$\ds \int f \cdot \chi_N \rd \mu = 0$

That is:

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

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.9 \, \text{(ii)}$