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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 :)

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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.

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