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A map into $R^{n}$ is measurable exactly when its coordinates are measurable

Statement

Let $n \geq 1$ , let $(X, A)$ be a measurable space, and let $f = (f_{1}, \dots, f_{n}) : X \to R^{n}$ . Then $f$ is measurable if and only if each coordinate function $f_{j} : X \to R$ is measurable.

Facts & Assumptions

Given: A natural number $n \geq 1$ , a measurable space $(X, A)$ , and a function $f = (f_{1}, \dots, f_{n}) : X \to R^{n}$ .

[L1]

Rational open boxes generate the Borel sigma-algebra on $R^{n}$ . (For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n)

[L2]

A generating family on the codomain suffices to test measurability. (A generating family on the codomain suffices to test measurability)

[L3]

Real-valued measurability is equivalent to threshold measurability. (Threshold characterisations of real-valued and extended-real-valued measurability)

Proof

technique · direct

1.1

Suppose $f$ is measurable. Fix $j$ and a real $a$ . Then

L3given

${x : f_{j} (x) > a} = f^{- 1} ⁣ (R^{j - 1} \times (a, \infty) \times R^{n - j}) .$

The displayed strip is open, hence Borel in $R^{n}$ , so the preimage is measurable. By [L3], each coordinate $f_{j}$ is measurable. [L3, given]

1.2

Conversely, suppose every coordinate $f_{j}$ is measurable. Let [L1, L3, algebra] $B = \prod_{j = 1}^{n} (a_{j}, b_{j})$ be a rational open box. Then

$f^{- 1} (B) = ⋂_{j = 1}^{n} {x : a_{j} < f_{j} (x) < b_{j}},$

and each factor on the right is measurable by [L3]. Therefore $f^{- 1} (B) \in A$ for every rational open box $B$ . [L1, L3, algebra]

2.1

By [L1] and [L2], step 1.2 implies that $f$ is measurable. Together with [step 1.1, step 1.2, L1, L2] step 1.1, this proves the equivalence.

step 1.1step 1.2L1L2∎

Depends on

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Sources

Sheldon Axler, Measure, Integration and Real Analysis, Section 2B (standard reference, not scraped)

A map into RnRn is measurable exactly when its coordinates are measurable