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


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.

  1. 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.

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

  1. Sums and Products of continuous functions are continuous
  2. Polynomials and (more generally) convergent power series are continuous on the circle of convergence.
  3. Compositions of continuous functions are continuous.
  4. 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,

  1. 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 $\pi_i ^{-1}((\alpha, \infty))$)).
  2. Products and sums of measurable functions are measurable.
  3. Continuous functions are measurable.
  4. Compositions of continuous functions are measurable.
  5. 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

  1. Pointwise limits of measurable functions are measurable.
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  • $\begingroup$ I can find all properties that you mention at my textbook, thank you for your answer $\endgroup$ – Ergin Suer Dec 18 '14 at 22:19
  • $\begingroup$ I tried to give a proof by using your answer. Could you check my answer? Here I didn't use the fact that the projections are continuous. Was your proof different with mine? Thanks. $\endgroup$ – Ergin Suer Dec 19 '14 at 20:58

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.

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  • $\begingroup$ You never defined $f$ and $g$. Why is it "clear" that $F_1$ is measurable? The rest seems nice. $\endgroup$ – PhoemueX Dec 19 '14 at 21:59
  • $\begingroup$ Sorry, I added the definitions of $f$ and $g$. In addition for any Borel set $B$ in $\mathbb{R}$, we have $F_1^{-1}(B)=A\times [0,\pi]$ for some $A\subseteq\mathbb{N}$. Since $\mu$ is the counting measure, all subsets of $\mathbb{N}$ are measurable. So $F_1^{-1}(B)$ is a measurable rectangle. $\endgroup$ – Ergin Suer Dec 19 '14 at 22:11
  • $\begingroup$ Ok, this makes it a fine proof. But I would (maybe) add the argument for $F_1$ in the answer. $\endgroup$ – PhoemueX Dec 19 '14 at 22:14

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