Continuity of a function of bounded variation

Question

Let $f: \mathbb{R} \rightarrow \mathbb{C}$ be a function of bounded variation. Then it is well known that it has right and left limits everywhere and only jump discontinuities. Set $\tilde{f}(x)=f(x+)$, the right limits of $f$. Suppose $\lambda\in\mathbb{R}$ is such that $\tilde{f}(\lambda+)=\tilde{f}(\lambda-)=\tilde{f}(\lambda)$ (i.e. a point of continuity for $\tilde{f}$). Is it also true that $\lambda$ is a point of continuity of $f$?

$EDIT:$ This is wrong. In fact, consider the function $\chi_{(-\infty,0) \cup (0,\infty)}$. It has bounded variation and its "right continuous" version is just $1$ everywhere, which is continuous at $0$ while the original function isn't. The problem lies in that I assumed somehow that functions of bounded variation only have jump discontinuities, which is wrong! They might have so-called removable discontinuities, where the limits from the left and right at a point agree but they do not agree with the value of the function at that point.

Answer 1

Addendum

There is a potential flaw in the original proof (below) which assumes that convergence to right-hand limits is uniform. Here is a different proof.

Any function of bounded variation has at most countably many jump discontinuities. There are two possibilities.

(1) There are finitely many jump discontinuities in some open neighborhood of $\lambda$. In this case, there is an interval $(\lambda - \delta, \lambda + \delta)$ where $f$ is continuous except possibly at $\lambda$. For $x \neq \lambda$ we have $f_R(x) = f(x+) = f_L(x) = f(x-) = f(x).$ By hypothesis $f_R$ is continuous at $\lambda$ and, therefore, $f_R(\lambda-) = f_R(\lambda+)$.

Thus $$f(\lambda-) = \lim_{x \to \lambda -}f(x) = \lim_{x \to \lambda -}f_R(x) = f_R(\lambda-) = f_R(\lambda+) = \lim_{x \to \lambda +}f_R(x) = \lim_{x \to \lambda +}f(x) = f(\lambda+),$$

and $f$ is continuous at $\lambda.$

(2) Jump discontinuities accumulate at $\lambda$. Let $(y_n)$ and $(z_n)$ be any increasing and decreasing sequences of points, respectively, both converging to $\lambda$. Since, the sums of the jumps must be bounded by the total variation of $f$, we have

$$\sum_{k=1}^\infty |f_R(y_k) - f_L(y_k)| < \infty, \\ \sum_{k=1}^\infty |f_R(z_k) - f_L(z_k)| < \infty,$$

and $\lim_{k \to \infty} |f_R(y_k) - f_L(y_k)| = \lim_{k \to \infty} |f_R(z_k) - f_L(z_k)| = 0$.

Thus,

$$f_L(\lambda-) = \lim_{k \to \infty} f_L(y_k) = \lim_{k \to \infty} f_R(y_k) - \lim_{k \to \infty} (f_R(y_k) - f_L(y_k)) = f_R(\lambda-), \\ f_L(\lambda+) = \lim_{k \to \infty} f_L(z_k) = \lim_{k \to \infty} f_R(z_k) - \lim_{k \to \infty} (f_R(z_k) - f_L(z_k)) = f_R(\lambda+), $$

and

$$f_L(\lambda-) = f_L(\lambda+) = f_R(\lambda+) = f_R(\lambda-).$$

Original Proof

By hypothesis, $\tilde{f}$ is continuous at $\lambda$. We must have $f(\lambda-) = f(\lambda+) =\tilde{f}(\lambda)$ when $ f$ is continuous at $\lambda$.

Suppose $f$ is not continuous at $\lambda$. This implies $f(\lambda-) \neq f(\lambda+) = \tilde{f} (\lambda).$ Since the left-hand limit exists there exists a sequence $(x_n)$ converging to $\lambda$ from the left such that $f(x_n)$ does not converge to $\tilde{f}(\lambda)$.

Hence, there exists $\epsilon_0 > 0$ such that for all sufficiently large $n$,

$$|f(x_n) - \tilde{f}(\lambda)| > \epsilon_0.$$

Now choose a sequence of points $(y_n)$ such that $x_n - 1/2^{n} < y_n < x_n.$ Then $y_n \to \lambda$ and $|f(x_n) - \tilde{f}(y_n)| < \epsilon_0/2$ for sufficiently large $n$, since the right-hand limit exists everywhere (and convergence is uniform in a compact neighborhood of $\lambda$).

Thus,

$$|\tilde{f}(y_n) - \tilde{f}(\lambda)| > |f(x_n) - \tilde{f}(\lambda)| - |f(x_n) - \tilde{f}(y_n)| > \epsilon_0/2, $$

which contradicts the continuity of $\tilde{f}$ at $\lambda.$

Answer 2

Following an idea of RRL, I wanted to prove this statement by using the Jordan decomposition, but I found what I think is a simpler way to do it.

We use the following lemma:

$\mathbf{Lemma}$ In the notation of the OP, we have that $\tilde{f}(\lambda+)=f(\lambda+)$ and $\tilde{f}(\lambda-)=f(\lambda-)$ .

$\mathit{Proof}$: Since $f$ only has a countable set of discontinuities, there exist$^{1}$ sequences $\Lambda_{n}$, $\lambda_{n}$ of points of continuity of $f$ such that $\Lambda_{n} \rightarrow \lambda+$ and $\lambda_{n}\rightarrow \lambda-$. By definition, on the set $C$ of points of continuity of $f$, $\tilde{f}|_{C}=f|_{C}$. Thus:

$\tilde{f}(\lambda+)=\lim_{n\rightarrow\infty}\tilde{f}(\Lambda_{n})=\lim_{n\rightarrow\infty} f(\Lambda_{n})=f(\lambda+)$

since both limits exist. We can work similarly using $\lambda_{n}$ to get the other claim:

$\tilde{f}(\lambda-)=\lim_{n\rightarrow\infty}\tilde{f}(\lambda_{n})=\lim_{n\rightarrow\infty} f(\lambda_{n})=f(\lambda-)$

$\blacksquare$

By the lemma, if for $\lambda \in \mathbb{R}$, $ \tilde{f}(\lambda+)=\tilde{f}(\lambda-)$, we also have $f(\lambda+)=f(\lambda-)$ and since $f$ is of bounded variation, it only has jump discontinuities, and thus $\lambda$ is a point of continuity for $f$.

What does this have to do with the Jordan decomposition? I simply noticed that this also works for increasing (or decreasing) functions since it really only needs that a) discontinuities form a countable set and b) there are only jump discontinuities.

1. This follows from the following argument: To find a sequence converging to $\lambda$ from the right, for each $n\in \mathbb{N}$ we pick a point of continuity $\Lambda_{n}$ such that $\Lambda_{n} \in (\lambda, \lambda+1/n)$. This can be done since each of these intervals does not consist only of points of discontinuity since then $f$ would have uncountably many discontinuities. By construction, $\Lambda_{n} \rightarrow \lambda+$. We can work similarly to get $\lambda_{n}$.