# If the Riemann-Liouville fractional integral of $f$ is zero then $f=0$ a.e

I have to prove this:

Let $$\alpha\in(0,1)$$ and $$f\in L^q(a,b)$$, $$1\leq q<\frac 1\alpha$$, and $$\mathcal{I}_{a+}^\alpha f=0$$. Then $$f(x)=0$$ for almost all $$x\in (a,b)$$.

Where $$(\mathcal{I}_{a+}^\alpha f)(x):=\frac{1}{\Gamma(\alpha)} \int_{a}^{x} f(y)(x-y)^{\alpha-1}$$.

Can Somebody help me? Thanks

$$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}$$Let $$g:=\Ga(\al)\mathcal{I}_{a+}^\al f$$, so that $$g=0$$ on $$(a,b)$$. Then for any $$z\in(a,b)$$ \begin{aligned} 0&=\int_a^z dx\,(z-x)^{-\al}g(x) \\ &=\int_a^z dx\,(z-x)^{-\al}\int_a^x dy\,f(y)(x-y)^{\al-1} \\ &=\int_a^z dy\,f(y)\int_y^z dx\,(z-x)^{-\al}(x-y)^{\al-1} \\ &=\int_a^z dy\,f(y)\int_0^1 du\,(1-u)^{-\al}u^{\al-1} \quad \Big[u:=\frac{x-y}{z-y}\Big] \\ &=\int_a^z dy\,f(y)\,\Ga(1-\al)\Ga(\al); \end{aligned} the interchange of the order integration is possible because $$f\in L^q(a,b)$$ for some $$q\ge1$$ and hence $$f\in L^1(a,b)$$.
So, $$\int_a^z f=0$$ for all $$z\in(a,b)$$ and thus $$f=0$$ almost everywhere (a.e.) on $$(a,b)$$. $$\quad\Box$$
Details on the last sentence in the proof above: Let $$\mu$$ be the measure over $$(a,b)$$ defined by the formula $$\mu(A):=\int_A f$$ for all Lebesgue-measurable $$A\subseteq(a,b)$$. Then the condition that $$\int_a^z f=0$$ for all $$z\in(a,b)$$ implies that $$\mu=0$$ on the algebra over $$(a,b)$$ generated by all intervals of the form $$(a,z)$$ for $$z\in(a,b)$$. By the uniqueness of the extension of measure, $$\mu=0$$ on all Borel subsets of $$(a,b)$$. Also, clearly $$\mu=0$$ on the sets of Lebesgue measure $$0$$. So, $$\mu=0$$ on all Lebesgue subsets of $$(a,b)$$. So, for each real $$t>0$$ and $$A_t:=f^{-1}((t,\infty))$$ we have $$0=\mu(A_t)\ge t|A_t|$$, where $$|\cdot|$$ is the Lebesgue measure. So, $$|A_t|=0$$ for all real $$t>0$$. So, $$f\le0$$ a.e. on $$(a,b)$$. Similarly, $$f\ge0$$ a.e. on $$(a,b)$$. Thus, $$f=0$$ a.e. on $$(a,b)$$.
The condition $$q<1/\al$$ was not needed or used in the proof above.