My question is -
Let, $f:[a,b]\to\Bbb{R}$ be a continuous function. Define $g:[a,b]\to\Bbb{R}$ by $$ g(x)= \begin{cases} f(a), & \text{when}\quad x=a\\ \operatorname{sup}_{(a,x]} f, & \text{otherwise} \end{cases} $$ Prove that $g$ is continuous on $[a,b]$.
I am able to prove the continuity of $g$ at $a$ in the following manner-
By continuity of $f$ at $a$, we get - for any $\varepsilon>0\exists\ \delta>0$such that $f(a)-\varepsilon<f(x)<f(a)+\varepsilon\ \forall x\in(a,a +\delta)$.
Thus for any $x\in(a,a +\delta)$, we have $f(a)-\varepsilon<\operatorname{sup}_{(a,x]} f<f(a)+\varepsilon\implies|g(x)-g(a)|<\varepsilon\ \forall x\in(a,a +\delta)$, since $f(a)=g(a)$, done.
Now, I want to work with the reamining $(a,b]$. So,
I take a point $x_0\in(a,b]$ and $f$ is continuous at $x_0$.
Choose $\varepsilon>0$. Then for this $\varepsilon>0,\exists\delta>0$ such that $f(x_0)-\varepsilon<f(x)<f(x_0)+\varepsilon\ \forall x\in(x_0 -\delta,x_0 -\delta)$.
My claim is - $g(x_0)-\varepsilon<g(x)<g(x_0)+\varepsilon\ \forall x\in(x_0 -\delta,x_0 -\delta)$ or equivalently $\operatorname{sup}_{(a,x_0]} f-\varepsilon<\operatorname{sup}_{(a,x]} f<\operatorname{sup}_{(a,x_0]} f+\varepsilon\ \forall x\in(x_0 -\delta,x_0 -\delta)$.
But I can't prove it rigorously.
Can anybody complete my proof? Thanks for assistance in advance.