While I was reading an article on Wikipedia which deals with pointwise convergence of a sequence of functions I asked myself how bad can the limit function be? When I say bad I mean how discontinuous it can be?
So I have these two questions:
1) Does there exist a sequence of continuous functions $f_n$ defined on the closed (or, if you like you can take open) interval $[a,b]$ (which has finite length) which converges pointwise to the limit function $f$ such that the limit function $f$ has infinite number of discontinuities?
2) Does there exist a sequence of continuous functions $f_n$ defined on the closed (or, if you like you can take open) interval $[a,b]$ (which has finite length) which converges pointwise to the limit function $f$ such that the limit function $f$ has infinite number of discontinuities and for every two points $c\in [a,b]$, $d\in [a,b]$ in which $f$ is continuous there exist point $e\in [c,d]$ in which $f$ is discontinuous?
I stumbled upon Egorov´s theorem which says, roughly, that pointwise convergence on some set implies uniform convergence on some smaller set and I know that uniform convergence on some set implies continuity of the limit function on that set but I do not know can these two questions be resolved only with Egorov´s theorem or with some of its modifications, so if someone can help me or point me in the right direction that would be nice.
Best Answer
The following is a standard application of Baire Category Theorem:
Set of continuity points of point wise limit of continuous functions from a Baire Space to a metric space is dense $G_\delta$ and hence can not be countable.
Another result is the following:
Any monotone function on a compact interval is a pointwise limit of continuous functions.
Such a function can have countably infinite set of discontinuities. For example in $[0,1]$ consider the distribution function of the measure that gives probability $1/2^n$ to $r_n$ where $(r_n)$ is any enumeration of rational numbers in $[0,1]$. The set of discontinuity points of this function is $\mathbb{Q}\cap[0,1]$.