A.E. Equal Positive Measurable Functions have Equal Integrals

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

Let $f, g: X \to \overline \R_{\ge 0}$ be positive $\mu$-measurable functions.

Suppose that $f = g$ almost everywhere.

Then:


 * $\displaystyle \int f \rd \mu = \int g \rd \mu$

Corollary
Let $f: X \to \overline \R$ be a $\mu$-integrable function, and $g: X \to \overline \R$ be measurable.

Suppose that $f = g$ almost everywhere.

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


 * $\displaystyle \int f \rd \mu = \int g \rd \mu$

Proof
Let $N$ be the set defined by:


 * $N = \set {x \in X: \map f x \ne \map g x}$

By hypothesis, $N$ is a $\mu$-null set.

If $N = \O$, then $f = g$, trivially implying the result.

If $N \ne \O$, then by Set with Relative Complement forms Partition:


 * $X = N \cup \paren {X \setminus N}$

Now:

which establishes the result.