Jensen's Inequality (Measure Theory)/Convex Functions

Theorem

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

Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.

Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a convex function.

Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:

$\map V {\dfrac {\ds \int g \cdot f \rd \mu} {\ds \int f \rd \mu} } \le \dfrac {\ds \int \paren {V \circ g} \cdot f \rd \mu} {\ds\int f \rd \mu}$

where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.

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Proof

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Let $\d \map \nu x := \dfrac {\map f x} {\ds \int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure.

This article, or a section of it, needs explaining.
In particular: This proof invokes a probability measure. Needs to be for a measure space. Does the proof work for both?
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Let $\ds x_0 := \int \map g s \rd \map \nu s$.

Then by convexity there exists constants $a, b$ such that:

$\ds \map V {x_0}$

$=$

$\ds a x_0 + b$

$\ds \forall x \in \R_{\ge 0}: \, $

$\ds \map V x$

$\ge$

$\ds a x + b$

In other words, there is a tangent line at $\tuple {x_0, V_0}$ that falls below the graph of $V$.

Therefore:

$\ds \map V {\map g s}$

$\ge$

$\ds a \map g s + b$

$\ds \leadsto \ \ $

$\ds \int \map V {\map g s} \rd \map \nu s$

$\ge$

$\ds a \int \map g s \rd \map \nu s + b$

Integration with respect to $\map \nu s$

$\ds $

$=$

$\ds \map V {x_0}$

$\ds $

$=$

$\ds \map V {\int \map g s \rd \map \nu s}$

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