# Showing a sequence of functions $f_n$ does not converge uniformly to $f$ on an interval.

Suppose for each $$n \in \mathbb{N}$$ we have a function $$f_n:[0,1] \to [0,1]$$ by $$f_n(x)=nx$$ on the interval $$x \in [0,\frac{1}{n}]$$ and $$1$$ if $$x \in (\frac{1}{n},1]$$, and define $$f=\lim_{n \to \infty} f_n$$.

I want to show that for any Lebesgue measurable $$B \subseteq [0,1]$$ with Lebesgue measure $$\lambda(B)=0$$ that the functions $$f_n$$ do not converge to $$f$$ uniformly on $$[0,1] \backslash B$$.

Now clearly we have the the limit $$f$$ is equal to $$0$$ at $$x=0$$ and is equal to $$1$$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $$B$$.

Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $$f_n$$ do not converge uniformly to $$f$$ as each of the functions $$f_n$$ is continuous we can use the uniform limit theorem to show that $$f_n$$ does not converge to $$f$$.

I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $$0$$, for example does continuity of each $$f_n$$ still hold in this case in which I can still apply the theorem?

Any insight would be much appreciated thanks :)

Take $$B \subset [0,1]$$ of measure zero and set $$l = \inf \ (0,1]\setminus B$$. If it were $$l > 0$$, then $$B$$ would contain some neighbourhood of $$0$$ (maybe without $$0$$ itself) which contradicts that $$|B| = 0$$. Hence $$l = 0$$ and thus we have a sequence $$x_n \to 0$$ contained in $$(0,1] \setminus B$$. Without loss of generality we can assume that $$0 < x_n \leq \frac{1}{2n}$$. Thus, $$f_n(x_n) \leq \frac{1}{2}$$ and in particular,
$$d_{[0,1] \setminus B}(f,f_n) \geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) \geq f(x_n) - \frac{1}{2} \stackrel{(x_n \neq 0)}{=} \frac{1}{2}.$$
Taking limits, we get that $$\lim_n d_{[0,1] \setminus B}(f,f_n) \geq \frac{1}{2} > 0$$ as desired.
Edit: note that we only used that $$B$$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.