A.E. Equal Positive Measurable Functions have Equal Integrals/Proof 2

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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:

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

From Measurable Functions Determine Measurable Sets, we have that:

$A$ is $\Sigma$-measurable.

Define $h : X \to \overline \R$ by:

$\map h x = \begin{cases}+\infty & x \in A \\ 0 & x \not \in A\end{cases}$

We can show that $h$ is $\Sigma$-measurable.

If $t < 0$, we have that:

$\set {x \in X : \map h x \le t} = \O$

From Sigma-Algebra Contains Empty Set, we therefore have that:

$\set {x \in X : \map h x \le t}$ is $\Sigma$-measurable if $t < 0$.

If $t \ge 0$, we have that:

$\set {x \in X : \map h x \le t} = A$

This is also $\Sigma$-measurable, so $h$ is a $\Sigma$-measurable function.

Clearly also $h \ge 0$.

We can show that:

$f \le g + h$

If $x \in A$, then $\map h x = +\infty$, so $\map g x + \map h x = +\infty$, and the inequality is satisfied trivially.

If $x \not \in A$, we have $\map f x = \map g x = \map g x + \map h x$, and so $\map f x \le \map g x + \map h x$ again.

From Integral of Positive Measurable Function is Monotone, we then have:

$\ds \int f \rd \mu \le \int \paren {g + h} \rd \mu$

Applying Integral of Integrable Function is Additive, we then have:

$\ds \int f \rd \mu \le \int g \rd \mu + \int h \rd \mu$

To compute the latter integral, note that:

$x \in X$ is such that $\map h x \ne 0$ if and only if $x \in A$.

Since by hypothesis, $f = g$ $\mu$-almost everywhere, we have that:

$\map \mu A = 0$

So:

$h = 0$ $\mu$-almost everywhere.

Then, by Integrable Function Zero A.E. iff Absolute Value has Zero Integral, we have:

$\ds \int \size h \rd \mu = 0$

Since $h \ge 0$, this gives:

$\ds \int h \rd \mu = 0$

and so:

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

Similarly, we have:

$g \le f + h$

and obtain:

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

simply swapping $g$ for $f$ in the prior computation.

We therefore have:

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

$\blacksquare$