Triangle Inequality for Integrals/Proof 2
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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 |
$\blacksquare$