Triangle Inequality for Integrals/Proof 2

Theorem

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

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

Then:

$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$

Proof

We have:

\(\ds \size {\int f \rd \mu}\)

\(=\)

\(\ds \size {\int f^+ \rd \mu - \int f^- \rd \mu}\)

Definition of Integral of Integrable Function

\(\ds \)

\(\le\)

\(\ds \size {\int f^+ \rd \mu} + \size {-\int f^- \rd \mu}\)

Triangle Inequality for Real Numbers, since $f$ is $\mu$-integrable both integrals are certainly real

\(\ds \)

\(=\)

\(\ds \int f^+ \rd \mu + \int f^- \rd \mu\)

\(\ds \)

\(=\)

\(\ds \int \paren {f^+ + f^-} \rd \mu\)

Integral of Positive Measurable Function is Additive

\(\ds \)

\(=\)

\(\ds \int \size f \rd \mu\)

Sum of Positive and Negative Parts

$\blacksquare$