An integrable function, but not necessarily a continuous one, is continuous when integrated.

Question

Let $f:\left[0,1\right]\to\mathbb{R}$ be an integrable function, but not necessarily continuous, with the property that $0\leq f(s)\leq 1$. Let $g:\left[0,1\right]\to\mathbb{R}$ be the function given by: $$ g(t)=\int_{0}^{t}f(s)\,ds. $$ Prove that $g$ is continuous.

Answer 1

We shall solve this by the $\epsilon-\delta$ definition.

Let $t>t_0$. Then, we have - $$|g(t)-g(t_0)| = \left|\int_{t_0}^tf(s)ds\right|$$ Now, as $0<f(s)<1$, $|f(s)|<1$, and we get - $$\left|\int_{t_0}^tf(s)ds\right| \le \int_{t_0}^t|f(s)|ds\le |t-t_0|$$ Thus, $|g(t)-g(t_0)|\le|t-t_0|$, and continuity follows by taking $\delta = \epsilon$ in the $\epsilon-\delta$ definition.

Answer 2

This proof is for Riemann integrable function, i.e. bounded.

$g(t_2)-g(t_1)=\int_{t_1}^{t_2}f(s)ds$. Since $|f(s)|\le M$, $|g(t_2)-g(t_1)|\le M(t_2-t_1)$. For any $\epsilon$, let $\delta=\frac{\epsilon}{M}$. So that for any $\epsilon \gt 0$, there is a $\delta$ where $|t_2-t_1|\lt \delta$, $|g(t_2)-g(t_1)|\lt \epsilon$.