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A Function into the the Product Sigma Algebra is Measureable iff its components are Measureable

Last updated Nov 1, 2022

# Statement

Let $(X, \mathcal{M})$, $(Y_{\alpha}, \mathcal{N}\alpha)$ be Measure Spaces for $\alpha \in A$. Let $Y = \prod\limits{\alpha \in A} Y_\alpha$ and let $\mathcal{N} = \bigotimes\limits_{\alpha \in A} \mathcal{N}\alpha$. Let $f : X \to Y$ and denote $f{\alpha} := \pi_{\alpha} \circ f$, where $\pi_\alpha$ is the Projection Map for $\alpha \in A$. Then $f$ is $(\mathcal{M}, \mathcal{N})$-measureable If and Only If $f_\alpha$ is $(\mathcal{M}, \mathcal{N}_\alpha)$-measureable for every $\alpha \in A$.

# Proof

$(\Rightarrow)$: If $f$ is a Measureable Function, then so is $f_{\alpha}$ since Projection Maps are Measureable Functions and Composition of Measureable Functions is Measureable. $\checkmark$

$(\Leftarrow)$: Recall that $$\bigotimes\limits_{\alpha \in A} \mathcal{N}{\alpha}= \sigma(\mathcal{E} :={\pi{\alpha}^{-1}(E) : \forall \alpha \in A, \forall E \in \mathcal{N}\alpha})$$ Then for every $E \in \mathcal{E}$, we have that $E = \prod\limits{\beta \in A} E_{\beta}$ where there is some $\alpha \in A$ so that $$E_{\beta} = \begin{cases}
E_{\alpha} \in \mathcal{N}{\alpha} & \text{ if } \alpha = \beta \\ Y{\beta} & \text{ otherwise } \end{cases}$$ for some $E_{\alpha} \in \mathcal{N}{\alpha}$. Then $$\begin{align*} f^{-1}(E) &= \bigcap\limits{\beta \in A} f_{\beta}^{-1}(E_{\beta})\\ &=f_{\alpha}^{-1}(E_\alpha)\\ &\in \mathcal{M}. \end{align*}$$ Since A Function is Measureable iff Preimages of Generating Sets are in the Source Sigma-Algebra, we have that $f$ is measureable. $\checkmark$ $\blacksquare$

# Remarks

  1. Indeed we can view the Product Sigma Algebra as the coarsest Sigma Algebra for which the Projection Maps are Measureable Functions.

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