Limit of integral of probability density function Is $\lim_{x \to 0} \int_{0}^{x}f(z) f_{A}(x-z)dz = 0$? Here $f(.)$, $f_{A}(.)$ are probability density functions. How do you prove this ?

This is true for any integrable function $f$.  Consider arbitrarily small $\varepsilon$ and the integral $\displaystyle\int_0^{\varepsilon}{fdx}$.  Then, assuming $f$ has a max $m$ the integral is bound by $$m\varepsilon\geq\int_{0}^{\varepsilon}{f(x)dx}$$ and by letting $\varepsilon\rightarrow0$ obviously the value of the integral goes to $0$.  This can be extended to $\mathbb{R}^n$ pretty simply.  Here you have multiple bounds, but considering the integral with fixed upper bounds $a_i$ together with a variable upper bound $\varepsilon$, the bound on the integral is $$m\varepsilon\prod_i{a_i}\geq\int_0^{a_1}\!\!\!\int_0^{a_2}\cdots\int_0^{a_{n-1}}\!\!\!\int_0^{\varepsilon}{f(x)dx}$$ and the bound can still be brought to $0$ by letting $\varepsilon\rightarrow 0$.  This is the same argument as the M-L inequality in complex analysis (http://en.wikipedia.org/wiki/Estimation_lemma), and I hope this helps.
