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Closure properties of measurable functions used by the integral

Statement

Let $(X, A, μ)$ be a measure space.

If $f, g : X \to \overline{R}$ are measurable and $f + g$ is defined pointwise, then $f + g$ is measurable.
If $c \in R$ and $f : X \to R$ is measurable, then $c f$ is measurable.
If $f : X \to R$ is measurable, then $f^{+} = \max {f, 0}$ , $f^{-} = \max {- f, 0}$ , and $∣ f ∣ = f^{+} + f^{-}$ are measurable.
If $E \in A$ and $f : X \to [0, + \infty]$ is measurable, then $f χ_{E}$ is measurable, where this function equals $f$ on $E$ and $0$ off $E$ (in particular, $0 \cdot (+ \infty) = 0$ here).
If $(f_{n})$ is a sequence of measurable functions $X \to \overline{R}$ , then $\inf_{n} f_{n}$ is measurable; if moreover $f_{n} ↑ f$ pointwise, then $f$ is measurable.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and functions or sets as in the relevant clause.

[L1]

A function $h : X \to \overline{R}$ is measurable exactly when ${h > a} \in A$ for every real $a$ (Extended-real-valued measurable functions).

[L2]

A sigma-algebra contains $X$ and $\emptyset$ and is closed under complements and countable unions; countable intersections follow by taking complements (Sigma-algebras).

[L3]

The rationals are countable and dense in the reals ( $Q$ is countably infinite, Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

Proof

technique · direct

1.1L1L2L3

For any extended-real-valued $h$ , the identities ${h < a} = ⋃_{q \in Q, q < a} (X ∖ {h > q})$ and ${h > a} = ⋃_{q \in Q, q > a} (X ∖ {h < q})$ follow from rational density, including when $h (x)$ is infinite. Thus [L2] and [L3] show that measurability of all strict sublevels is equivalent to measurability of all strict superlevels.

1.2givenL1L2

Put $h = f χ_{E}$ as defined in clause 4. For $a < 0$ , ${h > a} = X$ , since $h \geq 0$ . For $a \geq 0$ , ${h > a} = E \cap {f > a}$ . Both sets are measurable, including at $a = 0$ , so clause 4 follows.

1.3L1L2L3given

For a pointwise-defined sum, ${f + g > a} = ⋃_{q \in Q} ({f > q} \cap {g > a - q})$ . Indeed, if both summands are finite and their sum exceeds $a$ , choose a rational strictly between $a - g (x)$ and $f (x)$ . If one summand is $+ \infty$ , the other is not $- \infty$ , and a rational meeting the two inequalities still exists; if a summand is $- \infty$ , the defined sum cannot exceed $a$ . The reverse inclusion follows by adding the inequalities. The union is countable and measurable, proving clause 1.

2.1L1L2step 1.1

For $c > 0$ , ${c f > a} = {f > a / c}$ ; for $c < 0$ , ${c f > a} = {f < a / c}$ . If $c = 0$ , each superlevel is either $X$ or $\emptyset$ . Step 1.1 and [L1] therefore prove clause 2.

2.2givenL1L2step 1.1

Put $u = \inf_{n} f_{n}$ and $v = \sup_{n} f_{n}$ . The defining order properties of infimum and supremum give ${u < a} = ⋃_{n} {f_{n} < a}$ and ${v > a} = ⋃_{n} {f_{n} > a}$ , including infinite values. By step 1.1 and [L2] these sets are measurable, so $u$ and $v$ are measurable by [L1]. If $f_{n} ↑ f$ , then $f = v$ pointwise. This proves clause 5.

3.1L1L2step 1.3step 2.1

The function $- f$ is measurable by step 2.1. For any real-valued measurable $h$ , the superlevel of $\max (h, 0)$ is $X$ when $a < 0$ and ${h > a}$ when $a \geq 0$ . Apply this to $h = f$ and $h = - f$ to obtain measurable $f^{+}$ and $f^{-}$ . They are finite-valued, so step 1.3 makes $∣ f ∣ = f^{+} + f^{-}$ measurable, proving clause 3.

Closure properties of measurable functions used by the integral

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources