A.E. Equal Positive Measurable Functions have Equal Integrals/Corollary 2

Corollary to A.E. Equal Positive Measurable Functions have Equal Integrals
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g : X \to \overline \R$ be positive $\Sigma$-measurable functions.

Suppose that $f = g$ $\mu$-almost everywhere.

Let $A \in \Sigma$.

Then:


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

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


 * $\ds \int_A f \rd \mu = \int \paren {f \times \chi_A} \rd \mu$

and:


 * $\ds \int_A f \rd \mu = \int \paren {g \times \chi_A} \rd \mu$

We show that:


 * $f \times \chi_A = g \times \chi_A$ $\mu$-almost everywhere.

Since:


 * $f = g$ $\mu$-almost everywhere

there exists a $\mu$-null set $N \subseteq X$ such that:


 * whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$.

Suppose that $x \in X$ is such that:


 * $\map {\paren {f \times \chi_A} } x \ne \map {\paren {g \times \chi_A} } x$

From the definition of pointwise multiplication, we have:


 * $\map f x \map {\chi_A} x \ne \map g x \map {\chi_A} x$

So:


 * $\map f x \ne \map g x$

Hence:


 * $x \in N$

So:


 * $f \times \chi_A = g \times \chi_A$ $\mu$-almost everywhere.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we therefore have:


 * $\ds \int \paren {f \times \chi_A} \rd \mu = \int \paren {g \times \chi_A} \rd \mu$

and so:


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