Why the convergence is pointwise in this question and why the translation is continuous?

Question

I was reading this question here Proving that $f'$ is measurable on $\mathbb R$ if$f$ is differentiable on $\mathbb R$.

But I did not understand the following points in the solution:

1- why the translation $f_{h}(x)= f(x + h)$ continuous? could anyone give me a rigorous proof for this, please?

2- Why the convergence is pointwise everywhere here: " Now, you have $$\frac{f_h - f}{h} \to f'$$ pointwise everywhere"? could anyone give me a rigorous proof for this, please?

Answer 1

$f'(x) =\lim_{n \to \infty} f_n(x)$ where $f_n(x)=\frac {f(x+\frac 1 n)-f(x)} {1/n}$. So it is enough to check that each $f_n$ is measurable. But $f_n$ is continuous and that finishes the proof.

1) If $x_k \to x$ then $x_k+\frac 1 n \to x+\frac 1 n$ so $f(x_k+\frac 1 n) \to f(x+\frac 1 n)$ in view of the fact that any differentiable function is continuous.

2) This is by definition of derivative.

Answer 2

By definition, $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{f_h(x)-f(x)}{h}$$ And for 2, just check the definition of pointwise convergence.