Let $g$ be summable on $[a,b]$. Show that $f(x)=\int\_{[a,x]} g \, dm$ is continuous on $[a,b]$. ($m$ is Lebesgue measure on $\mathbb{R}$)

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First let $g$ be a simple function. For simplicity assume that
$$
 g(t) =
 \begin{cases} 
 \hfill y\_1 \hfill & \text{ if $t\in[a,c)$} \\
 \hfill y\_2 \hfill & \text{ if $t=c$} \\
\hfill y\_3 \hfill & \text{ if $t\in (c,b]$} \\
 \end{cases}
$$

Assume that $x\in [a,c)$. It is clear that $$\int\_{[a,x]} g \, dm=y\_1 m([a,x))=y\_1(x-a)$$

which is a linear function so is continuous. We can prove other cases similarly using fact that measure of a single point is $0$.

Is this correct? Also how can I generalize this to summable functions (not simple ones)?