Integral of Integrable Function over Measurable Set is Well-Defined

Theorem

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

Let $E \in \Sigma$.

Let $f: X \to \overline \R$ be a $\mu$-integrable function.

Then the $\mu$-integral of $f$ over $E$:

$\ds \int_E f \rd \mu = \int f \cdot \chi_E \rd \mu$

is well-defined.

Proof

We need to show that $f \cdot \chi_E$ is $\mu$-integrable, so that its $\mu$-integral is well-understood.

Since $f$ is $\mu$-integrable, it is certainly $\Sigma$-measurable.

From Characteristic Function Measurable iff Set Measurable, we have that:

$\chi_E$ is $\Sigma$-measurable.

Then by Pointwise Product of Measurable Functions is Measurable, we have:

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

We now want to verify that:

$\ds \int \paren {f \cdot \chi_E}^+ \rd \mu < \infty$

and:

$\ds \int \paren {f \cdot \chi_E}^- \rd \mu < \infty$

Note that from the definition of a characteristic function:

$\map {\chi_E} x \in \set {0, 1}$

so, for $x \in X$:

$\map {f \cdot \chi_E} x \ge 0$ if and only if $\map {\chi_E} x = 0$ or $\map {\chi_E} x = 1$ and $\map f x \ge 0$

and:

$\map {f \cdot \chi_E} x \le 0$ if and only if $\map {\chi_E} x = 0$ or $\map {\chi_E} x = 1$ and $\map f x \le 0$

So, we have:

$\paren {f \cdot \chi_E}^+ = f^+ \cdot \chi_E$

and:

$\paren {f \cdot \chi_E}^- = f^- \cdot \chi_E$

We can see that:

$f^+ \cdot \chi_E \le f^+$

and:

$f^- \cdot \chi_E \le f^-$

Since $f$ is $\mu$-integrable, we have:

$\ds \int f^+ \rd \mu < \infty$

and:

$\ds \int f^- \rd \mu < \infty$

Then, from Integral of Positive Measurable Function is Monotone, we have:

$\ds \int f^+ \cdot \chi_E \rd \mu \le \int f^+ \rd \mu < \infty$

and:

$\ds \int f^- \cdot \chi_E \rd \mu \le \int f^- \rd \mu < \infty$

This gives:

$\ds \int \paren {f \cdot \chi_E}^+ \rd \mu < \infty$

and:

$\ds \int \paren {f \cdot \chi_E}^- \rd \mu < \infty$

as required.

$\blacksquare$