Measurable Function Zero A.E. iff Absolute Value has Zero Integral/Corollary

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

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

Let $A, B \in \Sigma$ have $A \subseteq B$.

Then:


 * $\ds \int_A f \rd \mu = \int_B f \rd \mu$




 * $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.

Proof
We can write:


 * $B = A \cup \paren {B \setminus A}$

From Integral of Positive Measurable Function over Disjoint Union, we have:


 * $\ds \int_B f \rd \mu = \int_A f \rd \mu + \int_{B \setminus A} f \rd \mu$

Since:


 * $\ds \int_B f \rd \mu = \int_A f \rd \mu$

we get:


 * $\ds \int_{B \setminus A} f \rd \mu = 0$

From the definition of the $\mu$-integral over $A$, we have:


 * $\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu$

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


 * $\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu = 0$ $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.

So:


 * $\ds \int_A f \rd \mu = \int_B f \rd \mu$




 * $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.