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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.

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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, $$ \lim_{h\to 0^+} h^{-1} \int_0^h g(t)\,dt = g(0). $$

Does this answer your question?

EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.

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  • $\begingroup$ What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact. $\endgroup$ – I.C. Jan 7 at 22:00

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