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Every nonnegative measurable function is the increasing limit of simple measurable functions

Statement

Let $f : X \to [0, + \infty]$ be measurable. Then there is an increasing sequence of nonnegative simple measurable functions $(s_{n})$ such that $s_{n} (x) ↑ f (x)$ for every $x \in X$ .

Facts & Assumptions

Given: A measurable function $f : X \to [0, + \infty]$ .

[L1]

For a measurable $f$ , the sets $f^{- 1} (B)$ are measurable for every Borel $B \subseteq \overline{R}$ (Extended-real-valued measurable functions).

[L2]

A nonnegative measurable function with finite range is a nonnegative simple measurable function (Nonnegative simple measurable functions).

[L3]

Increasing pointwise suprema of measurable functions are measurable, and measurable functions remain measurable under the elementary truncations used below (Closure properties of measurable functions used by the integral).

Proof

technique · direct

1.1

Set $s_{0} : = 0$ . For $n \geq 1$ and $0 \leq k < n 2^{n}$ , put [L1, L2, construct] $E_{n, k} : = {x : k 2^{- n} \leq f (x) < (k + 1) 2^{- n}} \cap {f < n},$ and set $s_{n} : = \sum_{k = 0}^{n 2^{n} - 1} k 2^{- n} χ_{E_{n, k}} + n χ_{{f \geq n}} .$ Each $E_{n, k}$ and ${f \geq n}$ is measurable by [L1], the range of $s_{n}$ is finite, and therefore each $s_{n}$ is simple by [L2]; $s_{0}$ is also simple.

2.1

For each $x$ , one has $0 \leq s_{n} (x) \leq f (x)$ . If $f (x) < + \infty$ and [step 1.1, algebra] $n > f (x)$ , then $f (x) - 2^{- n} < s_{n} (x) \leq f (x)$ ; if $f (x) = + \infty$ , then $s_{n} (x) = n$ . Hence $s_{n} (x) \to f (x)$ .

3.1

The functions are increasing. Indeed, $s_{n} (x)$ is a dyadic multiple of [step 2.1, L3, algebra] ∎ $2^{- n}$ below $f (x) \land n$ , hence also a dyadic multiple of $2^{- (n + 1)}$ below $f (x) \land (n + 1)$ ; so the defining maximality of the $(n + 1)$ -st dyadic truncation gives $s_{n} (x) \leq s_{n + 1} (x)$ . Therefore $s_{n} ↑ f$ , in accord with [L3].

Depends on

Used by

Additivity of the nonnegative Lebesgue integral Corollary
Integrating against a Dirac measure is evaluation at the point Example
Integrable simple functions are dense in L¹(μ) Theorem
Integrating against a density agrees with integrating the product Theorem
Monotone convergence for the integral Theorem

Dependency tree · two levels

6 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 F. Bass, Real Analysis for Graduate Students, Theorem 7.1 setup (standard reference, not scraped)
John K. Hunter, Measure Theory Notes, Definition 4.4 (standard reference, not scraped)