# Proof that bounded right continuous functions are integrable.

I am reading Davie's book "One parameter Semigroups", on page 16 in the proof that "weak semigroups" are also "strong semigroups" it claims that for a right continuous locally bounded function $$g:\mathbb{R}\rightarrow\mathbb{R}$$, the integral

$$\varepsilon^{-1} \int_0^\varepsilon g(t) dt$$ converges.

I am assuming that since it is for $$\varepsilon \approx 0$$ then the same would be true for a general interval of integration $$[a,b]$$ if we assume $$g$$ bounded instead of locally bounded. If it is not, then the main question is why, under the hypotesis in bold text, that integral converges.

I am also assuming it refers to Riemman integrability.

Note: The book works with a specific function but as far as I am concerned the only hypothesis neccesary are those written above. In case the claim is not true, counterexample needed, then I would edit the question to include the particular functions used.

Since $$g$$ is right continuous at $$0$$, for every $$\varepsilon>0$$ there exists $$\delta>0$$ such that $$|g(y)-g(0)| \leq \varepsilon$$ when $$0\leq y\leq \delta$$.
Then in particular, for $$0\leq h\leq \delta$$, the triangle inequality implies $$\left| h^{-1}\int_0^h (g(t) - g(0))\,dt \right| \leq h^{-1} \int_0^h |g(t)-g(0)|\,dt \leq h^{-1}\int_0^h \varepsilon = \varepsilon.$$
This means that, under these hypotheses, $$\lim_{h\to 0^+} h^{-1} \int_0^h g(t)\,dt = g(0).$$