Product measurable functions

a) suppose $f,g:X\to[-\infty,\infty]$ are measurable. Prove the sets

$$\{x:f(x)<g(x)\}, \ \{x:f(x)=g(x)\}$$

are measurables.

I know that if $f,g$ measurables then $h:X\to{[-\infty,\infty]\times{[-\infty,\infty]}}$, with, $h(x)=(f(x),g(x))$, is measurable.

But
$$T=\{(x,y):x<y\}$$ is closed in $[-\infty,\infty]\times{[-\infty,\infty]}$ and

$$h^{-1}(T)=\{x:f(x)<g(x)\}$$.

Similarly $R=\{(x,y): x=y\}$ is closed, so $$h^{-1}(R)=\{x:f(x)=g(x)\}$$.

Thus, $h^{-1}(T),h^{-1}(R)$ are measurables. Is correct?

b) Prove the set of points at which a sequence of measurable real-valued functions converges (to a finite limit) is measurable.

• For b, use the Cauchy criterion. One can rewrite universal quantifiers as intersections and existential quantifiers as unions. To get you started, suppose $A_n = \{ x : |f_j(x) - f_k(x)| < 1/n \text{ for all sufficiently large } j,k \}$. Then the desired set is $\bigcap_{n=1}^\infty A_n$. Now you just need to continue breaking down $A_n$.
– Ian
Sep 4, 2014 at 2:19

A function is measurable if $f^{-1}(B)$ is measurable for all $B$ measurable, right? Now, do you agree that if $f$ and $g$ are measurable, so its difference is again measurable?

If you agree, let $h = f-g$, then note that

a) $$\{x: f(x)=g(x)\}=\{x: f(x)-g(x)=0\}= h^{-1}(0)$$ which is measurable because $h$ and $\{0\}$ are measurable ( I used the same word to two different concepts)

$$\{x: f(x)<g(x)\}=\{x: f(x)-g(x)<0\}= h^{-1}([-\infty,0))$$

and the same argument works here.

b) This is a little harder and the expression it isn't simple as above. The secret is look to the set $$\{x \in X: \exists \lim_nf_n(x)\}$$ as where the sequence $f_n(x)$ is a Cauchy sequence. So you can write

$$\{x \in X: \exists \lim_nf_n(x)\}= \bigcap_{k=1}^{\infty}\bigcup_{n_0=1}^{\infty}\bigcap_{n,m\ge n_0}\{x\in X: |f_n(x)-f_m(x)|\le \frac{1}{k}\}$$

The answer is that. Now, you have to convince yourself that the right side of the equality is exactly the definition of being a Cauchy sequence.

Hope it helps! ;)