When proving that a sequence of functions converges uniformly on compact sets in $\mathbb{R}$, it is sufficient to show uniform convergence on $[a,b]$

Question

I was proving a theorem in probability theory which was related to the characteristic function (hence the Fourier-analysis tag as well), and I showed that it converges uniformly on closed intervals, and I think that is sufficient, but I wanted to check. Thank you very much in advance!

Edit:

Thanks to FShrike, I know the answer, but just to expand on it a bit further: So one direction of the equivalence is easy, if it converges on compacts it converges on closed intervals, as they are compact. The other direction is by what FShrike said, assume we have convergence on closed intervals, if $K$ is compact, in $R^n$ this is equivalent to closed and bounded, so it’s included in a closed interval, so the uniform convergence will also hold on the subset $K\subset[a,b]$.

Answer 1

Yes. Any compact set in $\Bbb R$ is contained in some $[a,b]$ (for finite reals $a<b$) and if a sequence of functions converges uniformly on $[a,b]$ it converges uniformly on any subset of $[a,b]$.

Answer 2

Details: For $\emptyset\ne C\subset\Bbb R,$ let $\|f-f_n\|_C=\sup\{|f(x)-f_n(x)|:x\in C\}.$

Then $f_n\to f$ uniformly on $C$ iff $\lim_{n\to\infty}\|f-f_n\|_C=0.$

Now if $C\subset D\subset\Bbb R$ and if $f_n\to f$ uniformly on $D,$ then $f_n\to f$ uniformly on $C...$

... because $0\le \|f-f_n\|_C\le \|f-f_n\|_D$ for every $n$, and $\lim_{n\to\infty}\|f-f_n\|_D=0.$

In particular, if $C$ is compact then there exists $D=[a,b]$ with $C\subset D.$