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For $\alpha >0$, Markov's inequality gives the displayed bound, so $\mathbb{E}|Y_n{|}^{\alpha }\to 0$ implies convergence in probability to zero even when the first absolute moment does not tend to zero. For products $Y_n={\prod }_{j=1}^n{U}_j$ of independent and identically distributed factors, it suffices that $\mathbb{E}|U_1{|}^{\alpha }<1$. With the uniform distribution on $(0,b)$, the product of independent uniform random variables formula makes this $\alpha \log b<\log (1+\alpha )$. Since $\log (1+\alpha )/\alpha \to 1$ as $\alpha \downarrow 0$, such an $\alpha$ exists whenever $b<e$.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.