Right and left continuity of a function Suppose we want to show that a function $f$ with domain interval $[a,b]$ is right-continuous at $a$ and left-continuous at $b$.
If we also know that $f$ is increasing then is it sufficient to show that:
$$f(a)\geq\inf_{x>a}f(x)\quad\text{and}\quad f(b)\leq\sup_{x<b}f(x)?$$
I am not sure why that is true. Why does increasing matter in this case?
I'd appreciate any help or hint. Thank you.
 A: The claim is more general than what you write.
Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is increasing and satisfies $f(b)\leq\sup_{x<b}f(x)$ for some point $b$ in the domain of $f$.
Then, for each $\epsilon > 0$, we can find $\delta>0$ such that $f(b-\delta)+\epsilon\geq f(b)$ or, equivalently, $f(b)-f(b-\delta)\leq\epsilon$.
Since $f$ is increasing, it follows that $f(b)-f(b^{\prime})\leq\epsilon$ for all $b^{\prime}$ such that $b-b^{\prime}\leq\delta$, establishing left-continuity.
A: $\underset{x\rightarrow a^+}{\lim}f(x)$ is monotone decreasing so if it is bounded below it will converge.  It is bounded below by f(a) so it converges by monotone convergence theorem.  If $\underset{x\rightarrow a^+}{\lim}f(x) = L$ then $L = \underset{x\rightarrow a^+}{\liminf}f(x) \leq f(a)$, but by virtue of incre f, $\underset{x\rightarrow a^+}{\liminf}f(x)\geq f(a)\implies \underset{x\rightarrow a^+}{\liminf}f(x) = f(a) = L$. Thus $\underset{x\rightarrow a^+}{\lim}f(x) = f(a)$. Same argument applies to the other side.  
