If a sequence of functions $f_n$ on a common domain $D$ converges uniformly to $f$ with only jump discontinuities, then $f$ has only jump discontinuities.
I am trying to prove this statement, starting from the definition of uniform convergence: $\forall \epsilon > 0$, there is $N$ such that $\forall x \in D$, $\forall n > N, |f_n(x) - f(x)| < \epsilon$.
Also, WLOG suppose that all $f_n$ for $n > N$ has a jump discontinuity at $x_0$, i.e., $\lim_{x \to x_0^+}f(x) \neq \lim_{x \to x_0-}f(x)$ (otherwise, we could just take a bigger $N$ and avoid considering the discontinuity in when defining $f$).
I'm not entirely sure how to start proving this. I'm thinking of showing that the right limit of $f(x_0)$ is not equal to the left limit after manipulating the definition of uniform convergence? Should I do it by contradiction - assume $f$ has discontinuity of the second kind?
Firstly, let us clarify the definition of jump discontinuity. Let $a\in D$ be an interior point of $D$. We say that $f$ is jump discontiuous at $a$ if both left-hand limit $f(a-):=\lim_{x\rightarrow a-}f(x)$ and right-hand limt $f(a+):=\lim_{x\rightarrow a+}f(x)$ exist and finite but it is false that $f(a)=f(a-)=f(a+)$.
Let $a\in D$ be an interior point. We go to prove that either $f$ is continous at $a$ or $f$ is jump discontinuous at $a$. Hence, we only need to prove that $f(a-)$ and $f(a+)$ exist and are finite. Choose $\delta>0$ such that $(a-\delta,a+\delta)\subseteq D$. Firstly, we show that $f(a+)$ exists and is finite. Let $(x_{n})$ be an arbitrary sequence in $(a,a+\delta)$ such that $x_{n}\rightarrow a$. We go to show that $\lim_{n\rightarrow\infty}f(x_{n})$ exists by contradiction. Suppose the contrary that $\lim_{n\rightarrow\infty}f(x_{n})$ does not exist in $\mathbb{R}$, then there exists $\varepsilon_{0}>0$ and a subsequence $(x_{n_{k}})$ such that $\left|f(x_{n_{k+1}})-f(x_{n_{k}})\right|>\varepsilon_{0}$ for the following reason:
Recall that $\lim_{n}f(x_{n})$ converges and is finite iff $\forall\varepsilon>0\,\exists N\,\forall>N\left(|f(x_{n})-f(x_{N})|\leq\varepsilon\right)$. Therefore, $\lim_{n\rightarrow\infty}f(x_{n})$ does not exist in $\mathbb{R}$ iff $\exists\varepsilon>0\,\forall N\,\exists n>N\left(|f(x_{n})-f(x_{N})|>\varepsilon\right)$. Choose $\varepsilon_{0}>0$ such that $\forall N\,\exists n>N\left(|f(x_{n})-f(x_{N})|>\varepsilon_{0}\right)$. Put $N=n_{1}=1$, then there exists $n_{2}>n_{1}$ such that $|f(x_{n_{2}})-f(x_{n_{1}})|>\varepsilon_{0}$. Put $N=n_{2}$, we obtain $n_{3}>n_{2}$ such that $|f(x_{n_{3}})-f(x_{n_{2}})|>\varepsilon_{0}$. By recursion theorem, we obtain a subsequence $(x_{n_{k}})$ such that $\left|f(x_{n_{k+1}})-f(x_{n_{k}})\right|>\varepsilon_{0}$ for all $k$.
Since $f_{n}\rightarrow f$ uniformly, there exists $N$ such that $|f(x)-f_{n}(x)|<\varepsilon_{0}/10$ whenever $n\geq N$ and $x\in D$. In particular, we have that \begin{eqnarray*} & & \left|f(x_{n_{k+1}})-f(x_{n_{k}})\right|\\ & = & \left|f(x_{n_{k+1}})-f_{N}(x_{n_{k+1}})\right|+\left|f_{N}(x_{n_{k+1}})-f_{N}(x_{n_{k}})\right|+\left|f_{N}(x_{n_{k}})-f(x_{n_{k}})\right|\\ & < & \varepsilon_{0}/5+\left|f_{N}(x_{n_{k+1}})-f_{N}(x_{n_{k}})\right|. \end{eqnarray*} It follows that $\left|f_{N}(x_{n_{k+1}})-f_{N}(x_{n_{k}})\right|>\left|f(x_{n_{k+1}})-f(x_{n_{k}})\right|-\varepsilon_{0}/5>\frac{4}{5}\varepsilon_{0}$. By assumption, $f_{N}(a+)$ exists (and is finite), so letting $k\rightarrow\infty$ yields $0\geq\frac{4}{5}\varepsilon_{0}$, which is a contradiction.
Next, we show that the limit $\lim_{n\rightarrow\infty}f(x_{n})$ is independent of the choice of the sequence $(x_{n})$ in the following sense: If $(x_{n}')$ is another sequence in $(a,a+\delta)$ such that $x_{n}'\rightarrow a$, then $\lim_{n}f(x_{n})=\lim_{n}f(x_{n}')$. Let $(x_{n})$ and $(x_{n}')$ be such sequences. Denote $l=\lim_{n}f(x_{n})$ and $l'=\lim_{n}f(x_{n}')$. Define a sequence $(y_{n})$ by $$ y_{n}=\begin{cases} x_{\frac{n+1}{2}}, & \mbox{if }n\mbox{ is odd}\\ x'_{\frac{n}{2},} & \mbox{if }n\mbox{ is even} \end{cases}. $$ Hence, $(y_{n})=(x_{1},x_{1}',x_{2},x_{2}',\ldots)$. By the above, $\lim_{n}f(y_{n})$ exists and is finite. In particular, $\lim_{n}f(y_{2n+1})=\lim_{n}f(y_{2n})$, i.e., $l=l'$.
Now, we can conclude that $f(a+)=\lim_{n}f(x_{n})$ for any arbitrary sequence $(x_{n})$ in $(a,a+\delta)$ that converges to $a$.