# Finitely many discontinuities and uniform convergence

Suppose that $$\left\{f_n\right\}$$ converges uniformly to $$f$$, and that each $$f_n$$ has at most $$M$$ discontinuities, where $$M\in \mathbb{N}$$ is a fixed value. The $$f_n$$ don't need to be discontinuous at the same points.

QUESTION: Does it necessarily follow that $$f$$ has at most $$M$$ discontinuities?

This occurred to me while I was taking a walk, and I was wondering if the above statement was true or if there is a counterexample(possibly pathological).

Yes $f$ has at most $M$ discontinuities. If $f$ has a discontinuity at $p$, there must be $\epsilon > 0$ such that in any neighbourhood of $p$ there are points $x, y$ with $|f(x) - f(y)| > \epsilon$. If $f_n \to f$ uniformly, for sufficiently large $n$ we have $|f_n - f| < \epsilon/3$, and then the condition of the last sentence holds for $f_n$ with $\epsilon$ replaced by $\epsilon/3$, so $f_n$ is also discontinuous at $p$.

The answer Robert Israel gave is spot on. But because I'm still a novice (undergrad), I had a hard time understanding where he was getting at at first. So I'm going to elaborate on his answer. I just joined StackExchange, and this is my first time posting an answer. Constructive criticism is appreciated.

To answer the question, first we need to show that if $$f$$ is discontinuous at $$c$$, then after a certain term in the sequence $$(f_n)$$, all the later terms in the sequence are also discontinuous at $$c$$. I prove this below. But you can skip the proof and go straight to the explanation is you want.

PROOF: Suppose $$f$$ is discontinuous at $$c$$. Then by negating the definition of continuity, we know that there exists $$\epsilon>0$$ such that for all $$\delta >0$$, there exists an $$x$$ in the domain of $$f$$ such that $$|x-c|<\delta$$ and $$|f(x)-f(c)|> \epsilon$$.

Choose such an $$\epsilon$$ that meets the requirements of the above definition/negation.
Then $$\forall \delta>0, \exists x$$ such that $$|x-c|<\delta$$ and $$\epsilon>|f(x)-f(c)|=|f(x)-f_n(x)+f_n(x)-f_n(c)+f_n(c)-f(c)|$$ $$\leq |f(x)-f_n(x)|+|f_n(x)-f_n(c)|+|f_n(c)-f(c)|$$ $$\forall$$ $$n\in\mathbb{N}$$.

Because the sequence of functions $$(f_n)$$ converges uniformly to $$f$$, then we may choose $$\epsilon/3$$ such that $$|f_n(c)-f(c)|<\epsilon/3$$ and $$|f(x)-f_n(x)|<\epsilon/3$$ for all $$n\geq N$$ for some fixed number $$N\in \mathbb{N}$$.

Combining this new information with our earlier work, we see that we see that $$\forall \delta>0, \exists x$$ such that $$|x-c|<\delta$$ and $$\epsilon>|f(x)-f_n(x)|+|f_n(x)-f_n(c)|+|f_n(c)-f(c)|< \epsilon/3 +|f_n(x)-f_n(c)|+\epsilon/3$$, $$\forall n\geq N$$.

Finally, $$\forall \delta>0, \exists x$$ such that $$|x-c|<\delta$$ and $$\epsilon/3 >|f_n(x)-f_n(c)|$$, $$\forall n\geq N$$.

At any given point, $$f$$ is either continuous or discontinuous. There is no in-between. From the original question, we take it as a given that all the terms in $$(f_n)$$ are continuous everywhere in the domain of $$f$$, except at a maximum of $$M$$ points for each function in the sequence (where the points of discontinuity might vary in location). (For simplicity's sake, I think of the domains of $$f$$ and every $$f_n$$ to only be wherever $$(f_n)$$ converges uniformly.) Now because $$(f_n)$$ converges uniformly to $$f$$, at any points where all terms in $$(f_n)$$ are continuous, $$f$$ is continuous on those points also. But wherever $$f$$ is discontinuous, after a certain term in $$(f_n)$$, all the terms share a discontinuity at the same point, by the above proof. Therefore, the number of discontinuities of $$f$$ cannot exceed $$M$$.

• really nice explaination...just fix some typos in proof section where you have begin with ''$\epsilon >$'' it should be ''$\epsilon <$'' – Believer Sep 11 at 10:11