Show that $f(x)=\int_{[a,x]} g \, dm$ is continuous on $[a,b]$

Question

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 $\mathbb{R}$)

---

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)?

Answer 1

Hint: Try using the definition of continuity. You want to make the following inequality

$$\left|\int_{[x, x+t]} g \, dm\right| < \varepsilon$$

You know the integral is finite, so $g$ should be bounded on this compact interval. Thus you can take $|t| < \delta$ with $\delta$ small to make the value on the LHS to be less than $M \delta$, where $M$ is the bound.

Answer 2

$$\int_{[a,x]} g=\int_{[a,b]}g\cdot \chi_{[a,x]},$$ where $\chi$ is the characteristic function. Apply the dominated convergence theorem to conclude the statement for any sequence of $x_n \rightarrow x$. The result then follows.