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Jensen's integral inequality for a probability measure

Statement

Let $(X, A, P)$ be a probability space, let $f \in L^{1} (P)$ be real-valued, let $I \subseteq R$ be an interval containing $f (x)$ for almost every $x$ , and let $φ : I \to R$ be convex with $φ \circ f \in L^{1} (P)$ . Then $φ ⁣ (\int f d P) \leq \int φ (f) d P .$

Facts & Assumptions

Given: A probability space $(X, A, P)$ , a real-valued integrable $f$ , an interval $I$ containing its almost-everywhere range, and a convex $φ : I \to R$ with $φ \circ f \in L^{1} (P)$ .

[L1]

A probability measure is a measure with total mass $1$ (Probability measures and probability spaces).

[L2]

The Lebesgue integral is linear on $L^{1}$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[L3]

Every slope between the one-sided derivatives of a convex function yields a supporting line at an interior point (Every slope between the left and right derivatives of a convex function gives a supporting line).

[L4]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

Proof

technique · direct

1.1

Put $m : = \int f d P$ . If $m$ lies in the interior of $I$ , apply [L3] to obtain a supporting line $ℓ (x) = φ (m) + a (x - m)$ with $ℓ (x) \leq φ (x)$ on $I$ . Integrating and using [L1] and [L2] gives $\int φ (f) d P \geq \int ℓ (f) d P = φ (m) + a (\int f d P - m \int 1 d P) = φ (m) .$

L1L2L3givenalgebra

1.2

Suppose instead that $m$ is an endpoint of $I$ , say the left endpoint. Then [L1, L2, L4, given] $f - m \geq 0$ almost everywhere and $\int (f - m) d P = \int f d P - m \int 1 d P = 0$ by [L1] and [L2]. Therefore $f = m$ almost everywhere by [L4], so $\int φ (f) d P = φ (m) = φ ⁣ (\int f d P) .$ The right-endpoint case is identical.

2.1

Steps 1.1 and 1.2 cover the interior and endpoint cases, so Jensen's [step 1.1, step 1.2] ∎ inequality holds on the whole interval $I$ .

Depends on

Used by

Jensen's inequality can fail on an infinite measure space without normalization Counterexample
Jensen's inequality yields the weighted AM-GM inequality Example
FALSE: Jensen's inequality holds on an infinite measure space without normalization False statement

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem (7.44) (standard reference, not scraped)