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Consider $g: I \rightarrow \mathbb{R}$ given by $$g(x) = \int_{x_0}^x f(y)dy$$ If $f : I \rightarrow \mathbb{R}$ is Riemann-integrable, then $g$ is continuous on $I$.

Proof: For any $x_1 \in I$, consider $$|g(x) - g(x_1)| = |\int_{x_0}^x f(y)dy - \int_{x_0}^{x_1} f(y)dy | = |\int_{x_1}^x f(y)dy | \le \int_{x_1}^x |f(y)|dy$$ If $x_1$ is not on the boundary of $I$, then there is a $\rho$ so $|x - x_1| < \rho \Rightarrow x \in I$. Since $f(x)$ is bounded for all such $x$(since it's Riemann-integrable on $(x_1 - \rho, x_1 + \rho))$, we know that in this interval $|f(x)| \le K$ for some $K$. If $x_1 < x$, then $|\int_{x_1}^x |f(y)|dy \le \int_{x_1}^x K dy = K(x - x_1)$. So, when $x \rightarrow x_1$, $|g(x) - g(x_1)| \rightarrow 0$, thus continuity.

Now, two questions: What if $x_1 > x$? And what if $x_1$ lies on the boundary of $I$? Especially the latter, I'm not sure how to argue?

john Sackson
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  • If $x_1>x$, then there should be no problem with the proof you have since you are only concerned with the absolute value of things (and so switching the limits of integration won't affect anything). No matter where $x_1$ is, since $f$ is Riemann-integrable on $I$, there should be a uniform bound on $f$. If you say "(since it's RI on $I$)" you should be able to get away with $x_1$ on the boundary. – user31415926535 Mar 15 '15 at 16:07

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