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I have to show that $h$ is measurable as well as $\int h d(\mu \times \nu) < \infty$ .

I tried showing by contradiction that $\int h$ had to be finite but I’m stuck with showing how it is measurable.

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**1- Measurability:**

**Fact:** a function is measurable if and only if it is the pointwise limit of a sequence of simple functions.

So take $s_n, t_n$ simple such that $f(x)=\lim s_n(x)$ for every $x$ and $g(y)=\lim t_n(y)$ for every $y$. Then $h(x,y)=f(x)g(y)=\lim s_n(x)t_n(y)$ for every $(x,y)$. It only remains to observe that each $(x,y)\longmapsto s_n(x)t_n(y)$ is a simple function to conclude that $h$ is measurable. This follows readily from

$$

1_A(x)1_B(y)=1_{A\times B}(x,y).

$$

**2- Integrability:**

We will use the monotone convergence theorem, asuming you have not proved Fubini yet. Otherwise, this is trivial. But how can you have Fubini if you don’t have measurability?

Take two nondecreasing sequences of nonnegative simple functions $s_n(x)$, $t_n(y)$ converging pointwise to $|f(x)|$ and $|g(y)|$ respectively. Then $s_n(x)t_n(y)$ is a nondecreasing sequence of nonnegative simple functions converging pointwise to $|h|$. By the monotone convergence theorem

$$

\int |h| \;d(\mu\times \nu)=\lim\int s_n(x)t_n(y)\;d(\mu\times \nu)(x,y).

$$

By definition of the product measure

$$

\int 1_{A\times B}\;d(\mu\times \nu)=(\mu\times \nu) (A\times B)=\mu(A)\nu (B)=\int 1_Ad\mu\int 1_Bd\nu.

$$

By linearity, this extends to simple functions. Hence, by monotone convergence again,

$$

\int s_n(x)t_n(y)\;d(\mu\times \nu)(x,y)=\int s_nd\mu\int t_nd\nu\longrightarrow \int |f|d\mu\int |g|d\mu<\infty.

$$

So $h$ is integrable with

$$

\int |h| \;d(\mu\times \nu)=\int |f|d\mu\int |g|d\nu.

$$

Note that applying the above to $f_{\pm}$ and $g_{\pm}$, we can deduce

$$

\int h \;d(\mu\times \nu)=\int fd\mu\int gd\nu.

$$

Hint: show that $\int f(x)g(y) d(\mu\times \nu)=\int f(x)(\int g(y)d\nu)d\mu$.

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