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A nonnegative measurable function with finite integral is finite almost everywhere

Statement

If $f : X \to [0, + \infty]$ is measurable and $\int f d μ < + \infty$ , then $f (x) < + \infty$ for almost every $x$ .

Facts & Assumptions

Given: A nonnegative measurable function $f$ with finite integral.

[L1]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L2]

The nonnegative integral of the simple function $n χ_{F}$ equals its simple integral $n μ (F)$ (The nonnegative integral agrees with the simple integral on simple functions).

[L3]

The real numbers form a complete ordered field, and for every real $x$ there is a positive integer $m > x$ (The Cauchy-sequence reals have the least-upper-bound property, The reals form a totally ordered field, Every complete ordered field is Archimedean).

Proof

technique · direct

1.1L1L2given

Let $F : = {f = + \infty}$ , which is measurable. For every positive integer $n$ , $n χ_{F} \leq f$ . By [L1] and [L2], $n μ (F) = \int n χ_{F} d μ \leq \int f d μ < + \infty$ .

2.1step 1.1L3algebrachoose∎

Put $J = \int f d μ < + \infty$ . If $μ (F) = + \infty$ , step 1.1 fails already for $n = 1$ . If $0 < μ (F) < + \infty$ , choose a positive integer $n > J / μ (F)$ by [L3]. Multiplication by the positive finite number $μ (F)$ gives $n μ (F) > J$ , contradicting step 1.1. Thus $μ (F) = 0$ , so the exceptional set where $f$ is infinite is null. Equivalently, $f < + \infty$ almost everywhere.

A nonnegative measurable function with finite integral is finite almost everywhere

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources