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all! I got stuck on this question today, although it seemed straight forward when I started. Here is the proposed problem:

Let a and b be extended real numbers with a $\lt$ b. Prove that if f is a bounded, monotone function on the interval (a, b), then $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ both exist and are finite.

I started by using the fact that f has a sup and inf, meaning that the limits must exist and be finite because x $\epsilon$ (a,b). I am just not sure how to prove this rigorously. Thank you!

rlh282
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2 Answers2

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The idea is that monotonicity implies the inf and sup must be the one-sided limits at the end points.

Suppose $f$ is monotonically increasing. Pick $\epsilon > 0$. Let $\sup_{x \in (a,b)}f(x) = M$. Since $M$ is a least upper bound, there exists some $c \in (a,b)$ such that $M - f(c) < \epsilon$. Let $\delta = b - c$. Then if $b - x < \delta$ for $x \in (a,b)$, we have $x > c$, hence $f(x) \geq f(c)$ by monotonicity. Therefore $M - f(x) \leq M - f(c) < \epsilon$.

Ethan Alwaise
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Without loss of generality, suppose that $f$ is increasing. Let $L = \sup\{f(x) : x \in (a,b)\} < \infty$. We wish to show that $\lim_{x \to a^-}f(x) = L$. That is, for any $\epsilon > 0$, there is a $\delta$ such that $a-\delta < x < a \implies |f(x) - L| < \epsilon$.

To that end, let $\delta$ be such that $f(a - \delta) > L - \epsilon$, noting that such a $\delta$ must exist by the definition of a supremum. Now, to fill in the blanks, show that this $\delta$ satisfies the above requirement.

Ben Grossmann
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