Suppose $f_n\rightarrow f$ pointwise. Show that if each $f_n$ is continuous then $f$ is also continuous.

Question

I know that this result is true for uniform convergence, but I'm struggling to find a counter example of such function.

Answer 1

This is false: for instance (classic example), take $f_n\colon[0,1]\to\mathbb{R}$ defined by $f_n(x) = x^n$. What is the pointwise limit $f$?

Answer 2

This is false:

let $f_n(x) = \left\{ \begin{array}{r l} nx, &x\in[0,\frac 1 n]\\ 1, &x\in[\frac 1 n, 1] \end{array} \right.$

Answer 3

To add to the answers above, note that the counterexamples, while not continuous everywhere, are still continuous on dense subsets of domain. This is in fact always the case: if $f_n: X \to \mathbb{R}$ is a family of functions, and $f_n \to f$ pointwise, and all $f_n$ are continuous, then even though $f$ is not necessarily continuous everywhere, it must be continuous on a dense subset of the domain.

I believe the assumptions on the domain necessary for this conclusion are metrizable and locally compact, as Baire theorem is used in the proof.