If $f$ is increasing real valued function and $g$ is defined as following then $g$ is increasing and right continuous.

Let $f:[a,b]\mapsto\mathbb{R}$ be increasing function. Now define, $g:[a,b]\mapsto\mathbb{R}$ such that $g(x)= f(x+)$ for all $x\in [a,b)$ and $g(b)=f(b)$. Now prove that $g$ is increasing function and $g$ is right continuous.

$f(x+)= inf\{f(t) | x<t\leq b\}$ for all $x\in[a,b)$. Since $f$ is increasing function so $f(x+)$ exists for all $x\in[a,b)$.

Now since for $x<y$ so $\{f(t) | y<t\leq b\}\subset\{f(t)| x<t\leq b\}$. Therefore $g(x)\leq g(y)$.

But I am not sure about the right continuous.

What I need, for $y\in [a,b)$ I need to show that for all $\epsilon>0$ there exists $\delta>0$ such that if $y< x< y +\delta$ implies that $| g(x) - g(y) | < \epsilon$ and since $g$ is increasing function so $g(y)\leq g(x)$.