How to prove the measurability of a function? Isn't this very obvious? What kind of proof is expected for this kind of question?

The definition of measurability says a mapping $f: E \rightarrow F$ is said to be measurable relative to $\mathcal{E}$ and $\mathcal{F}$ if $f^{-1} B \in \mathcal{E}$ for every $B$ in $\mathcal{F}$.

So here, $f: E\rightarrow F$ is measurable relative to $\mathcal{E}$ and $\mathcal{F}$, so $f^{-1}B \in \mathcal{E}$ for every $B$ in $\mathcal{F}$ and $g: E \rightarrow G$ is measurable relative to $\mathcal{E}$ and $\mathcal{G}$ so $g^{-1}C \in \mathcal E$ for every C in $G$. Hence, $h^{-1}D = (f^{-1}D, g^{-1}D) \in \mathcal{E}$ for every D in $\mathcal{F} \times \mathcal{G}$.

It seems very wrong because I am just regurgitating the definitions.

• What's $(f^{-1}D,g^{-1}D)$? That doesn't look like a subset of $E$ in the first place. – Saucy O'Path Jun 30 '18 at 9:25

The product $\sigma$-algebra $\mathcal{F} \otimes \mathcal{G}$ by definition is generated by the family

$\mathcal{P}:= \{A \times B: A \in \mathcal{F}, B \in \mathcal{G}\}$

and so $h:(E, \mathcal{E}) \to (F \times G, \mathcal{F} \otimes \mathcal{G})$ is measurable iff

$$\forall A \times B \in \mathcal{P}: h^{-1}[A \times B] \in \mathcal{E}$$

This is a general fact on generating sets of $\sigma$-algebras (analogous to such properties for bases and subbases in topology):

If $f: (E,\mathcal{E}) \to (F, \mathcal{F})$ is a function between measurable spaces and $\mathcal{F} = \sigma(\mathcal{P})$, so the image $\sigma$-algebra is generated by some subfamily $\mathcal{P}$, then $f$ is measurable iff $\forall P \in \mathcal{P}: f^{-1}[P] \in \mathcal{E}$.

The proof is not hard: the left to right implication of the iff is the definition of measurability, and for the right to left implication, define $\mathcal{F}' = \{A \subseteq F: f^{-1}[A] \in \mathcal{E}\}$, and do the easy check that $\mathcal{F}'$ is a $\sigma$-algebra on $F$, using properties like $f^{-1}[F\setminus A] = E\setminus f^{-1}[A]$ and $f^{-1}[\bigcup_n A_n] = \bigcup_n f^{-1}[A_n]$. By the right hand side assumption, $\mathcal{P} \subseteq \mathcal{F}'$ and so $$\mathcal{E} = \sigma(\mathcal{P}) \subseteq \mathcal{F}'$$ by minimality. But this exactly says that $f$ is measurable.

Finally, note that $h^{-1}[A \times B] = f^{-1}[A] \cap g^{-1}[B]$ and as $f^{-1}[A], g^{-1}[B] \in \mathcal{E}$ (because $f,g$ are measurable), so is their intersection. QED.

Let $p_F:F\times G\to F$ be prescribed by $(a,b)\mapsto a$.

Let $p_G:F\times G\to G$ be prescribed by $(a,b)\mapsto b$.

Then by definition of $\mathcal F\otimes\mathcal G$:

$h:(E,\mathcal E)\to(F\times G,\mathcal F\otimes\mathcal G)$ is measurable if and only if the functions $p_F\circ H:(E,\mathcal E)\to(F,\mathcal F)$ and $p_G\circ H:(E,\mathcal E)\to(G,\mathcal G)$ are both measurable.

So you are asked to prove that: $$p_F\circ H:(E,\mathcal E)\to(F,\mathcal F)\text{ and }p_G\circ H:(E,\mathcal E)\to(G,\mathcal G)$$ are indeed measurable.

"$f:(A,\mathcal A)\to(B,\mathcal B)$ is measurable" means in my answer that $f$ is a function $A\to B$ measurable wrt $\mathcal A$ and $\mathcal B$.

• Why the if and only if condition? I cannot see why they are equivalent – user10024395 Jun 30 '18 at 9:56
• $\mathcal F\otimes\mathcal G$ is by definition the smallest $\sigma$-algebra on $F\times G$ such that $p_F$ and $p_G$ are measurable. The composition of measurable functions is measurable, so if $H$ is measurable then also $p_F\circ H$ and $p_G\circ H$. Do you have troubles with the other side? – drhab Jun 30 '18 at 10:02
• The other side: if $p_{F}\circ H$ and $p_{G}\circ H$ are measurable then $H^{-1}\left(p_{F}^{-1}\left(\mathcal{F}\right)\cup p_{G}^{-1}\left(\mathcal{G}\right)\right)\subseteq\mathcal{E}$ and consequently $\sigma\left(H^{-1}\left(p_{F}^{-1}\left(\mathcal{F}\right)\cup p_{G}^{-1}\left(\mathcal{G}\right)\right)\right)\subseteq\mathcal{E}$ . Finally in general it can be shown that $\sigma\left(H^{-1}(\mathcal{V})\right)=H^{-1}(\sigma(\mathcal{V}))$ and applying that here gives $H^{-1}(\mathcal{F}\otimes\mathcal{G})\subseteq\mathcal{E}$ (i.e. $H$ is measurable). – drhab Jun 30 '18 at 10:23
• For a proof of $\sigma(H^{-1}(\mathcal V)=H^{-1}(\sigma(\mathcal V))$ see here. – drhab Jun 30 '18 at 10:30

While it is obvious that $h^{-1}[R\times S]=f^{-1}[R]\cap g^{-1}[S]\in\mathcal E$ for all $R\in\mathcal F$ and $S\in\mathcal G$, the proof must account for the fact that not all the sets of $\mathcal F\otimes\mathcal G$ are products of sets and, in point of fact, there aren't very explicit descriptions of a $\sigma$-algebra in terms of its generators.

The tool to use is the fact that if $\mathcal H$ is a $\sigma$-algebra on $H$ and $r:X\to H$ is a function, then $\{r^{-1}[K]\,:\, K\in \mathcal H\}$ is a $\sigma$-algebra as well.

Let $p_F$ and $p_G$ denote projections onto $F$ and $G$ respectively.

We have

\begin{align} \mathcal{F} \otimes \mathcal{G} &= \sigma(p_F^{-1}(\mathcal{F}) \cup p_G^{-1}(\mathcal{G})) \\ &= \sigma(\{p_F^{-1}(A) : A \in \mathcal{F}\} \cup \{p_G^{-1}(B) : B \in \mathcal{G}\}) \\ &= \sigma(\{A \times G : A \in \mathcal F\} \cup \{F \times B : B \in \mathcal G\})\\ &= \sigma(\{A \times B : A \in \mathcal F, B \in \mathcal G\}) \end{align}

whereas the last equality holds because $A \times B = (A \times G) \cap (F \times B)$.

Now for every $A \in \mathcal F, B \in \mathcal G$ we have $$h^{-1}(A \times B) = \underbrace{f^{-1}(A)}_{\in \mathcal{E}} \cap \underbrace{g^{-1}(B)}_{\in \mathcal{E}} \in \mathcal{E}$$

because $f$ and $g$ are measurable.

Verify that $\{C \subseteq F \times G : h^{-1}(C) \in \mathcal{E}\}$ is a $\sigma$-algebra on $E$ and clearly $$\{A \times B : A \in \mathcal F, B \in \mathcal G\} \subseteq \{C \subseteq F \times G : h^{-1}(C) \in \mathcal{E}\}$$

Then also $$\mathcal{F} \otimes \mathcal{G} = \sigma(\{A \times B : A \in \mathcal F, B \in \mathcal G\}) \subseteq \{C \subseteq F \times G : h^{-1}(C) \in \mathcal{E}\}$$

or in other words $h^{-1}(C) \in \mathcal{E}$ for every $C \in \mathcal{F} \otimes \mathcal{G}$. Hence $h$ is measurable w.r.t $\mathcal{E}$ and $\mathcal{F} \otimes \mathcal{G}$.