# 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:

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

$\blacksquare$