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

## Theorem

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

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

Let $g: X \to \overline \R$ be $\Sigma$-measurable.

Suppose that $f = g$ almost everywhere.

Then $g$ is also $\mu$-integrable, and:

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

## Proof

From Function Measurable iff Positive and Negative Parts Measurable, we have that:

$g^+$, $f^+$, $g^-$ and $f^-$ are all $\Sigma$-measurable.

From Functions A.E. Equal iff Positive and Negative Parts A.E. Equal, we have that:

$f^+ = g^+$ and $f^- = g^-$ $\mu$-almost everywhere.

Since $f^+$ and $g^+$ are positive $\Sigma$-measurable functions that are equal $\mu$-almost everywhere, we have:

$\ds \int f^+ \rd \mu = \int g^+ \rd \mu$

Similarly, we have that $f^-$ and $g^-$ are positive $\Sigma$-measurable functions that are equal $\mu$-almost everywhere, and so:

$\ds \int f^- \rd \mu = \int g^- \rd \mu$

Since $f$ is $\mu$-integrable, we have that:

$\ds \int f^+ \rd \mu < \infty$

and:

$\ds \int f^- \rd \mu < \infty$

Hence:

$\ds \int g^+ \rd \mu < \infty$

and:

$\ds \int g^- \rd \mu < \infty$

So $g$ is $\mu$-integrable.

We also have:

 $\ds \int g \rd \mu$ $=$ $\ds \int g^+ \rd \mu - \int g^- \rd \mu$ Definition of Integral of Integrable Function $\ds$ $=$ $\ds \int f^+ \rd \mu - \int f^- \rd \mu$ $\ds$ $=$ $\ds \int f \rd \mu$ Definition of Integral of Integrable Function

$\blacksquare$