prove that $F$ is $\mu\times\mathcal{L}$ measurable where $F(n,x)=\frac{(2n+1)^2\sin((2n+1)x)}{(n(n+1))^2}$ Let $\mu$ be the counting measure on $\mathbb{N}$ and $\mathcal{L}$ be the Lebesgue measure on $[0,\pi]$. Define the function $F$ on $\mathbb{N}\times\mathcal{L}$ by $F(n,x)=\frac{(2n+1)^2\sin((2n+1)x)}{(n(n+1))^2}$ 
Prove that $F$ is $\mu\times\mathcal{L}$ measurable. We need to show that for any $\alpha\in\mathbb{R}$ , $F^{-1}(\alpha,\infty)$ is a measurable set i.e it is in the $\sigma$-algebra generated by measurable rectangles. But I cannot determine $F^{-1}(\alpha,\infty)$? Thanks for your help

Edit:
Here is my second attempt: For every $\alpha \in \mathbb{R}$,   $F^{-1}(\alpha,\infty)$ is open in $\mathbb{N}\times [0,\pi]$ i.e it is a Borel measurable set. 


*

*Is it also $\mu\times \mathcal{L}$-measurable set?


Wiki says every Borel measurable set is a Lebesgue measurable set. So $F^{-1}(\alpha,\infty)$ is a Lebesgue measurable set. 


*Does it mean that  $F^{-1}(\alpha,\infty)$ is $\mathcal{L}\times\mathcal{L}$-measurable?

*If the answer of 2 is yes, can we conclude that $F^{-1}(\alpha,\infty)$ is $\mu\times\mathcal{L}$-measurable?

 A: Here is an important thing to keep in mind:
If you want to show that the function 
$$
\Bbb{R} \to \Bbb{R}, x \mapsto e^{x^2 \cdot \sin(x)} - \sum_{n=0}^\infty \frac{x^n}{(n!)^2}
$$
is continuous, you will (hopefully) not try to find for every $\varepsilon > 0$ some $\delta > 0$ such that ...
Instead, you will rely on certain closure properties of the class of continuous functions. I.e.


*

*Sums and Products of continuous functions are continuous

*Polynomials and (more generally) convergent power series are continuous on the circle of convergence.

*Compositions of continuous functions are continuous.

*Uniformly convergent sequences of continuous functions yield continuous limits.


Likewise, if you are trying to show that a "nice looking" function is measurable, you will normally not try to show that $f^{-1} ( (\alpha, \infty))$ is measurable for all $\alpha$. Instead, you will rely on certain closure properties of the class of measurable functions.
In this case,


*

*The projections $\pi_1 : \mathbb{N} \times \Bbb{R} \to \Bbb{R}, (n,x) \mapsto n$ and $\pi_2 : \mathbb{N} \times \Bbb{R} \to \Bbb{R}, (n,x) \mapsto x$ are measurable (if you do not know this, show it using the definition (i.e. calculate $\pi_i ^{-1}((\alpha, \infty))$)).

*Products and sums of measurable functions are measurable.

*Continuous functions are measurable.

*Compositions of continuous functions are measurable.

*If $f$ is measurable with $f(x) \neq 0$ for all $x$, then $1/f$ is measurable.


These properties (do you know them?) should allow you to show that $F$ is indeed measurable.
EDIT: I forgot to mention one of the most important closure properties of the class of measurable functions (because it was not needed in this case), namely


*Pointwise limits of measurable functions are measurable.

A: Let $F_1(n,x):=\frac{(2n+1)^2}{(n(n+1))^2}$ and $F_2(n,x):=sin((2n+1)x)$. So it is enough to show that $F_1$ and $F_2$ are $\mu\times\mathcal{L}$-measurable. It is clear that $F_1$ is $\mu\times\mathcal{L}$-measurable. For the measurability of $F_2$, we consider the maps $f: [0,\pi]\to \mathbb{R}$ by $f(x)=x$ and $g:\mathbb{N}\to\mathbb{R}$ by $g(n)=2n+1$. Clearly $f$ is $\mathcal{L}$-measurable and $g$ is $\mu$-measurable. By using this fact, the function $h:\mathbb{N}\times[0,\pi]\to\mathbb{R}$ defined by $h(n,x)=g(n)f(x)$ is$ \mu\times\mathcal{L}$-measurable. We know that $sin$ function is continuous, so it is Borel-measurable. So it is $\mathcal{L}$-measurable since every Borel-measurable function is also $\mathcal{L}$-measurable. So the composition $sin\circ h=F_2$ is $\mu\times\mathcal{L}$-measurable.
