A.E. Equal Positive Measurable Functions have Equal Integrals

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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:

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


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:

\(\displaystyle \int f \rd \mu\) \(=\) \(\displaystyle \int f \chi_X \rd \mu\) Characteristic Function of Universe
\(\displaystyle \) \(=\) \(\displaystyle \int f \chi_{N \cup \paren {X \setminus N} } \rd \mu\)
\(\displaystyle \) \(=\) \(\displaystyle \int f \paren {\chi_N + \chi_{X \setminus N} } \rd \mu\) Characteristic Function of Union
\(\displaystyle \) \(=\) \(\displaystyle \int f \chi_N \rd \mu + \int f \chi_{X \setminus N} \rd \mu\) Integral of Integrable Function is Additive
\(\displaystyle \) \(=\) \(\displaystyle 0 + \int f \chi_{X \setminus N} \rd \mu\) Integral of Integrable Function over Null Set
\(\displaystyle \) \(=\) \(\displaystyle \int g \chi_{X \setminus N} \rd \mu\) Definition of $N$
\(\displaystyle \) \(=\) \(\displaystyle \int g \chi_N \rd \mu + \int g \chi_{X \setminus N} \rd \mu\) Integral of Integrable Function over Null Set
\(\displaystyle \) \(=\) \(\displaystyle \int g \paren {\chi_N + \chi_{X \setminus N} } \rd \mu\) Integral of Integrable Function is Additive
\(\displaystyle \) \(=\) \(\displaystyle \int g \chi_X \rd \mu\) Characteristic Function of Union
\(\displaystyle \) \(=\) \(\displaystyle \int g \rd \mu\) Characteristic Function of Universe

which establishes the result.

$\blacksquare$


Sources