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 $M$.

  • $\begingroup$ really nice explaination...just fix some typos in proof section where you have begin with ''$\epsilon >$'' it should be ''$\epsilon <$'' $\endgroup$ – Believer Sep 11 '19 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.