Triangle Inequality for Integrals

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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:

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


Corollary

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

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


Then:

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


Real Number Line

On the real number line, the Triangle Inequality for Integrals takes the following form:

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.


Then:

$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$


Complex Plane

In the complex plane, the Triangle Inequality for Integrals takes the following form:

Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \C$ be a continuous complex function.


Then:

$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$

where the first integral is a complex Riemann integral, and the second integral is a definite real integral.


Complex Function

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

Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.

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


Then $\cmod f$ is $\mu$-integrable and:

$\ds \cmod {\int f \rd \mu} \le \int \cmod f \rd \mu$


Proof

We have:

\(\ds \size {\int f \rd \mu}\) \(=\) \(\ds \size {\int f^+ \rd \mu - \int f^- \rd \mu}\) Definition of Integral of Measure-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

$\blacksquare$


Also see


Sources

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