Product of improper integrable riemann function and integrable function.

Question

I have the following problem while working with linear differential equations:

I'm told to proof that the following system:

$t^{-\sigma}x' = A(t) x$ where $A: \mathbb{R} \to {M_N(\mathbb R)}$ is continuous and $\sigma \in (0,1)$

At some point I must verify that the following Picard's iterations are well defined and continuous:

$x_0(t) = x_0$

$x_{n+1}(t) = x_0+\int_{0}^ts^{-\sigma}A(s)x_n(s)ds$

So assuming $x_n$ is continuous I have to proof that: $s^{-\sigma}A(s)x(s)$ is integrable.

So the problem is given an improper integrable riemann function and a function that certainly is integrable (A(t)x(t) is continuous) is the product integrable?

It seems to me that this cannot be solved using Lebesgue caracterization of Riemann integrable functions as $t^{-\sigma}$ is not bounded.

Answer 1

Assuming that $x_n$ and $A$ are both continuous in a neighborhood of $0$, (you can prove this inductively for $x_n$), then $\|A(s)x_n(s)\|$ is bounded near $0$. Therefore $$\left\|\int_0^t s^{-\sigma}A(s)x_n(s)\,ds\right\| \le \int_0^t s^{-\sigma}\|A(s)x_n(s)\|\,ds \le M\int_0^t s^{-\sigma}\,ds$$ for some $M > 0$. Hence the function is Riemann Integrable for $\sigma < 1$.