# A.E. Equal Positive Measurable Functions have Equal Integrals/Proof 1

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## 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:

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

## 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:

\(\ds \int f \rd \mu\) | \(=\) | \(\ds \int f \chi_X \rd \mu\) | Characteristic Function of Universe | |||||||||||

\(\ds \) | \(=\) | \(\ds \int f \chi_{N \cup \paren {X \setminus N} } \rd \mu\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int \map f {\chi_N + \chi_{X \setminus N} } \rd \mu\) | Characteristic Function of Union | |||||||||||

\(\ds \) | \(=\) | \(\ds \int f \chi_N \rd \mu + \int f \chi_{X \setminus N} \rd \mu\) | Integral of Integrable Function is Additive | |||||||||||

\(\ds \) | \(=\) | \(\ds 0 + \int f \chi_{X \setminus N} \rd \mu\) | Integral of Integrable Function over Null Set | |||||||||||

\(\ds \) | \(=\) | \(\ds \int g \chi_{X \setminus N} \rd \mu\) | Definition of $N$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \int g \chi_N \rd \mu + \int g \chi_{X \setminus N} \rd \mu\) | Integral of Integrable Function over Null Set | |||||||||||

\(\ds \) | \(=\) | \(\ds \int \map g {\chi_N + \chi_{X \setminus N} } \rd \mu\) | Integral of Integrable Function is Additive | |||||||||||

\(\ds \) | \(=\) | \(\ds \int g \chi_X \rd \mu\) | Characteristic Function of Union | |||||||||||

\(\ds \) | \(=\) | \(\ds \int g \rd \mu\) | Characteristic Function of Universe |

which establishes the result.

$\blacksquare$