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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

1.1L1L2construct

Set $s_{0} : = 0$ . For $n \geq 1$ and $0 \leq k < n 2^{n}$ , put $E_{n, k} : = {x : k 2^{- n} \leq f (x) < (k + 1) 2^{- n}}$ 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.

2.1step 1.1algebra

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

3.1step 1.1step 2.1algebra∎

The functions are increasing. Indeed, $s_{n} (x)$ is the largest multiple of $2^{- n}$ at most $f (x) \land n$ , hence is also a multiple of $2^{- (n + 1)}$ at most $f (x) \land (n + 1)$ . The maximality of the latter dyadic truncation gives $s_{n} (x) \leq s_{n + 1} (x)$ . Together with step 2.1, this proves $s_{n} ↑ f$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources