One-Sided Continuity¶

Definition¶

Let $X \subseteq \mathbb{R}$ and let $f \colon X \to \mathbb{R}$ be a function. Let $x_0 \in X\text{.}$

Assume that there exists an $r > 0$ such that $f$ is defined for every $x \in [x_0, x_0 + r)\text{.}$ We say that $f$ is right-continuous at $x = x_0$ if for every $\varepsilon > 0$ there exists a number $\delta > 0$ with $0 < \delta \le r$ such that

x_0\le x<x_0+\delta \implies|f(x)-f(x_0)|<\varepsilon.

Assume that there exists an $r > 0$ such that $f$ is defined for every $x \in (x_0 - r, x_0]\text{.}$ We say that $f$ is left-continuous at $x = x_0$ if for every $\varepsilon > 0$ there exists a number $\delta > 0$ with $0 < \delta \le r$ such that

x_0-\delta<x\le x_0 \implies|f(x)-f(x_0)|<\varepsilon.

Interpretation¶

One-sided continuity asks for continuity at $x_0$ using the points on one side only: $f$ is right-continuous at $x_0$ if $f(x)$ is close to $f(x_0)$ for all $x\ge x_0$ close to $x_0$ , and left-continuous if the same holds for $x\le x_0$ $x_{0}$ . Equivalently, the one-sided limit from that side exists and equals $f(x_0)$ .

A function with a jump at x0: the left part approaches a value L, while the right part starts at the filled point f(x0), so the function is right-continuous but not left-continuous there.

The function in the figure jumps at $x_0$ . From the right the values start at the filled point $f(x_0)$ and stay close to it, so $\lim_{x\to x_0^+}f(x)=f(x_0)$ and $f$ is right-continuous at $x_0$ . From the left the values approach the lower height $L$ at the open circle, so $f$ is not left-continuous at $x_0$ , and therefore not continuous there.

The comparison is always with the value $f(x_0)$ itself, so including the point $x=x_0$ in $x_0\le x<x_0+\delta$ costs nothing: there the inequality reads $0<\varepsilon$ $ε$ . Moving the filled point in the figure down to height $L$ would make $f$ left-continuous instead of right-continuous: which side is continuous depends on where the value at the jump is placed.

One-sided continuity is what remains of continuity at an endpoint of an interval: a function on $[a,b]$ is continuous at $a$ exactly when it is right-continuous there.

A common slip is to read "right-continuous at $x_0$ " as "continuous on an interval to the right of $x_0$ ". The definition concerns the single point $x_0$ : a function can be continuous on $(x_0,x_0+r)$ and still fail to be right-continuous at $x_0$ .

Right-continuity returns in probability theory: every cumulative distribution function is right-continuous.

Key Results¶

Right-continuity and left-continuity are defined using one-sided limits.
If $f$ is defined on a neighbourhood $(x_0-r,x_0+r)$ of $x_0\text{,}$ then $f$ is continuous at $x_0$ if and only if it is both right-continuous and left-continuous at $x_0\text{.}$ At an endpoint of the domain, continuity coincides with the one-sided continuity that is available there.

Examples¶

The floor function $\lfloor x\rfloor$ (the largest integer $\leq x$ $x$ ) is right-continuous but not left-continuous at every integer; the ceiling function $\lceil x\rceil$ (the smallest integer $\geq x$ $x$ ) is left-continuous but not right-continuous there.

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 5

Aliases: right-continuous, left-continuous, continuous from the right, continuous from the left Keywords: right-continuity, left-continuity Symbols: delta, element of, epsilon, real numbers, subset

Last updated: 22 October 2026 19:04