$m(x) := \inf\{f(t): t \in [a, x)\}$ and $M(x) := \sup\{f(t): t \in [a, x)\}$ . Prove $m$ and $M$ are continuous from left on $(a,b)$

Question

Let $f$ be a bounded function on $[a,b]$. Define $m(x) = \inf\{f(t): t \in [a, x)\}$ and $M(x) = \sup\{f(t): t \in [a, x)\}$ . Prove that $m$ and $M$ are continuous from left on $(a,b)$

Could anyone help me please in solving this?

Answer 1

Let $\{x_n\}$ increase to $x$. If $\epsilon >0$ then there exists $y \in [a,x)$ such that $f(y) < m(x)+\epsilon$. Since $y <x$ we have $y<x_n$ for $n$ sufficiently large so $m(x_n) \leq f(y)< m(x)+\epsilon$ for $n$ sufficiently large. Also, $m(x) \leq m(x_n)$ for all $n$. This proves that $m(x_n) \to m(x)$. The argument for $M(x)$ is similar. You can also prove left continuity of $M$ using that of $m$ by changing $f$ to $-f$.