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If $f:\mathbb{R}\to\mathbb{R}$ is a left continuous function can the set of discontinuous points of $f$ have positive Lebesgue measure?

I wondered this today, but made little progress. Thank you.

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2 Answers 2

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There's nothing wrong with Ben Derrett's answer, but I just thought I'd point out the same argument goes through under slightly weaker assumptions.

Lemma. Let $f: \mathbb{R} \to \mathbb{R}$ be arbitrary. Then the set $A \subset \mathbb{R}$ of all points $a$ such that

  1. $f$ is discontinuous at $a$
  2. the limit of $f(x)$ as $x$ approaches $a$ from the left exists

is countable.

Proof. Suppose for contradiction that $A$ is uncountable. For $i \in \mathbb{Z}^+$, let $A_i$ be the set of all $a \in A$ such that, for any neighbourhood $U$ of $a$, $|f(x) - f(y)| \geq 1/i$ holds for some $x,y \in U$. Since $f$ is discontinuous on $A$, we have $A = \bigcup_{i=1}^\infty A_i$ and it follows that $A_n$ is uncountable for some $n$.

Call a point $a \in A_n$ left-isolated if $(a - \epsilon,a)$ does not meet $A_n$ for sufficiently small $\epsilon>0$. By associating to each left isolated point $a$ a rational number from the interval $(a-\epsilon,a)$, we see that $A_n$ has countably many left-isolated points. Since $A_n$ is uncountable, there exists some $p \in A_n$ which is not left isolated.

But now, given $\epsilon > 0$, we have that $(p - \epsilon, p)$ is a neighbourhood of some $a \in A_n$. Hence, $|f(x) - f(y)| \geq 1/n$ for some $x,y \in (p - \epsilon, p)$. But this contradicts the assumption that the limit as $x \to p$ from the left of $f(x)$ exists. QED.

As an application, after reproving the lemma with left-hand limits replaced by right-hand limits, we get the following.

Theorem. If $f$ is such that for all $a \in \mathbb{R}$ one of the limits $$ \lim_{x \to a^-} f(x) \text{ or } \lim_{x \to a^+} f(x) $$ exists, then $f$ has a countable number of discontinuities.

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Sam Watson gave me the following answer:

$f$ can have at most countably many discontinuous points.

Write $$S_n = \left\{x\in\mathbb{R}:\max\left(f(x)-\liminf_{y\downarrow x}~f(y),\limsup_{y\downarrow x}~f(y)-f(x)\right)>\frac{1}{n}\right\}$$

Then the set of discontinuities of $f$ is given by $\cup_{n\in\mathbb{N}}S_n$, so it suffices to prove that $S_n$ is countable for each $n$, since a countable union of countable sets is countable.

Fix $n\in\mathbb{N}$. Suppose that we have a bounded increasing sequence $x_i$ in $S_n$, with limit $x$. Then we can pass to a subsequence $x'_i$, such that for all $i$

$$f(x'_i)-\liminf_{y\downarrow x'_i}~f(y)>\frac{1}{n}$$

or that for all $i$

$$\limsup_{y\downarrow x'_i}~f(y)-f(x'_i)>\frac{1}{n}$$

Such a sequence contradicts left continuity at $x$, so any increasing sequence in $S_n$ must be unbounded, so $S_n$ is countable.

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    $\begingroup$ Why does the converging subsequence contradicts the left continuity? $\endgroup$
    – Joe_Chao
    Commented Nov 17, 2020 at 23:12
  • $\begingroup$ To see why an uncountable subset of $\mathbb{R}$ must contain an increasing sequence, notice that if we start with an interval that contains an uncountable set of discontinuities and consider its dyadic partition, there must be a step at which both the subintervals contain an uncountable set of discontinuities. $\endgroup$
    – No-one
    Commented May 3, 2022 at 10:24

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