A.E. Equal Positive Measurable Functions have Equal Integrals/Corollary 1

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:


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

from A.E. Equal Positive Measurable Functions have Equal Integrals.

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: