# Mapping between Euclidean Spaces Measurable iff Components Measurable

## Theorem

Let $\R^n$ and $\R^m$ be Euclidean spaces.

Denote by $\mathcal{B}^n$ and $\mathcal{B}^m$ their respective Borel $\sigma$-algebras.

Denote with $\mathcal B$ the Borel $\sigma$-algebra on $\R$.

Let $f: \R^n \to \R^m$ be a mapping, and write:

$f \left({\mathbf x}\right) = \begin{bmatrix}f_1 \left({\mathbf x}\right) \\ \vdots \\ f_m \left({\mathbf x}\right)\end{bmatrix}$

with, for $1 \le i \le m$, $f_i: \R^n \to \R$.

$\forall i:f_i: \R^n \to \R$ is $\mathcal{B}^n \, / \, \mathcal B$-measurable

## Proof

### Necessary Condition

Suppose that $f$ is $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable.

It is to be shown that for $1 \le i \le m$, $f_i: \R^n \to \R$ is $\mathcal{B}^n \, / \, \mathcal B$-measurable.

By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that:

$f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) \in \mathcal{B}^n$

for every open real interval $\left({a \,.\,.\, b}\right)$.

Since $f \left({\mathbf x}\right) = \mathbf y$ iff, for all $i$, $f_i \left({\mathbf x}\right) = y_i$, it follows that:

$\mathbf x \in f^{-1} \left({\mathbf y}\right) \iff \forall i: \mathbf x \in f_i^{-1} \left({y_i}\right)$

i.e., iff $\mathbf x \in \displaystyle \bigcap_{i \mathop = 1}^m f_i^{-1} \left({y_i}\right)$.

By Preimage of Union under Mapping, it follows that for any $B \in \mathcal{B}^m$:

$\mathbf x \in f^{-1} \left({B}\right) \iff \exists \mathbf y \in B: \forall i: \mathbf x \in f_i^{-1} \left({y_i}\right)$

In particular, consider the open set $B = \left\{{\mathbf y \in \R^n: y_i \in \left({a \,.\,.\, b}\right)}\right\}$.

Then by construction of $B$, $\mathbf x \in f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right)$ iff $f \left({x}\right) \in B$.

Therefore, $f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) = f^{-1} \left({B}\right)$.

Since $B$ is open, it is also measurable by Characterization of Euclidean Borel Sigma-Algebra.

Thus $f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) \in \mathcal{B}^n$ as $f$ was assumed $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable.

$\Box$

### Sufficient Condition

Suppose that all $f_i$ are $\mathcal{B}^n \, / \, \mathcal B$-measurable.

It is to be shown that $f$ is $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable.

By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that:

$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) \in \mathcal{B}^n$

for every half-open $m$-rectangle $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$.

Now observe, for all $\mathbf x \in \R^n$:

 $\displaystyle f \left({\mathbf x}\right)$ $\in$ $\displaystyle \left[ [{\mathbf a \,.\,.\, \mathbf b}\right))$ $\displaystyle \iff \ \$ $\displaystyle \forall i: f_i \left({\mathbf x}\right)$ $\in$ $\displaystyle \left[{a_i \,.\,.\, b_i}\right)$ $\displaystyle \iff \ \$ $\displaystyle \mathbf x$ $\in$ $\displaystyle \bigcap_{i \mathop = 1}^m f_i^{-1} \left[{a_i \,.\,.\, b_i}\right)$

Thus:

$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) = \displaystyle \bigcap_{i \mathop = 1}^m f_i^{-1} \left[{a_i \,.\,.\, b_i}\right)$

Now the $f_i^{-1} \left[{a_i \,.\,.\, b_i}\right)$ are measurable sets since the $f_i$ are $\mathcal{B}^n \, / \, \mathcal B$-measurable.

Since $\mathcal{B}^n$ is a $\sigma$-algebra, Sigma-Algebra Closed under Countable Intersection applies, and it follows that:

$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) \in \mathcal{B}^n$

$\blacksquare$