Dirichlet's test for convergence of improper integrals

Dirichlet's test claims that for two continuous functions $f,g\in[a,\infty]$ where $f,g\geq 0$, if a certain $M$ exists such that $\left|\int_a^bf(x)dx\right|\leq M$ for every $a\leq b$, and $g(x)$ is monotonically decreasing, and $\lim_{X\to\infty}g(x)=0$, then $\int_a^\infty fg$ is convergent.

Does this also apply for a non-continuous $f(x)$? $g(x)$ is still continuous.

This question relates to another question of mine, regarding a specific integral problem. If this is true, then my other problem will be solved.

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Your version of Dirichlet's test seem to be wrong: take $f\equiv 1$ and $g(x) = 1/x$. – Dirk May 4 '12 at 18:36
@Ory: Your condition should read there exists $M$ such that $\int_a^x f(t) dt \leq M$ for all $x > a$ – user17762 May 4 '12 at 18:40
@Ory: Also the restriction that $f$ is continuous might be a bit too strong. I think it should still work if $f$ is continuous except on a measure zero set. Hence, it will work for the problem you have posed in the other thread. – user17762 May 4 '12 at 18:41
@Marvis I've stated that such $M$ should exists by stating that $f(x)$ is upper-bounded. Will fix the question text, thanks. – Ory Band May 5 '12 at 9:41
Check this reference books.google.it/… – Siminore May 5 '12 at 10:09

I paste here a theorem from Mathematical Analysis by S. C. Malik,Savita Aror (see also Google Books here)

Theorem. If $\phi$ is bounded and monotonic in $[a,+\infty)$ and tends to zero at $+\infty$, and $\int_a^X f$ is bounded for $X \geq a$, then $\int_a^\infty f \phi$ is convergent.

As you can see, no continuity is really necessary. Honestly, the continuity assumption appears often for the sake of simplicity. Improper integrals can be defined as limits of Riemann integrals: all you need is local integrability. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught.

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