Integral of Positive Measurable Function over Measurable Set is Well-Defined

Theorem

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

Let $A \in \Sigma$.

Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.

Then the $\mu$-integral of $f$ over $A$ defined by:

$\ds \int_A f \rd \mu = \int \paren {\chi_A \cdot f} \rd \mu$

is well-defined.

Proof

We simply need to show that:

$\chi_A \cdot f$ is a positive $\Sigma$-measurable function.

For $x \in A$, we have:

\(\ds \map {\paren {\chi_A \cdot f} } x\)

\(=\)

\(\ds \map {\chi_A} x \map f x\)

Definition of Pointwise Multiplication

\(\ds \)

\(=\)

\(\ds \map f x\)

Definition of Characteristic Function of Set

\(\ds \)

\(\ge\)

\(\ds 0\)

For $x \in X \setminus A$, we have:

\(\ds \map {\paren {\chi_A \cdot f} } x\)

\(=\)

\(\ds \map {\chi_A} x \map f x\)

Definition of Pointwise Multiplication

\(\ds \)

\(=\)

\(\ds 0\)

Definition of Characteristic Function of Set

so:

$\chi_A \cdot f$ is non-negative.

We now show that $\chi_A \cdot f$ is $\Sigma$-measurable.

From Characteristic Function Measurable iff Set Measurable, we have:

$\chi_A$ is $\Sigma$-measurable.

From Pointwise Product of Measurable Functions is Measurable, we then have:

$\chi_A \cdot f$ is $\Sigma$-measurable.

So:

$\chi_A \cdot f$ is a positive $\Sigma$-measurable function.

So its $\mu$-integral is well-defined.

$\blacksquare$