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Let $f:[0,1]\to\mathbb{R}$ be continuous. Prove $$ \lim_{n\to\infty}\int_0^1\int_0^1\cdots\int_0^1f\left((x_1x_2\cdots x_n)^{1/n}\right)dx_1dx_2\cdots dx_n = f\left(\frac1e\right) $$ My idea so far is to use uniform distributions to calculate this. Let $X_1,X_2,\dots$ be independent Uniform$(0,1)$ variables. My idea is to somehow use that this integral is equal to $$ \lim_{n\to\infty}\mathbb{E}\left((X_1X_2\cdots X_n)^{1/n}\right). $$ I think I can somehow use the fact that $\frac1n\sum_{i=1}^n\log X_i\overset{a.s.}{\to}\mathbb{E}\log X_i=-1$. I think I have some of the ideas, but my question is how can I actually put together these ideas in a rigorous way prove this? Do I need the full power of the SLLN, or would I somehow be able to just use the WLLN? I also think maybe I will have to use the dominated convergence theorem, but I'm not sure. I'm struggling with self doubt on this question. Any help appreciated!

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  • $\begingroup$ You can indeed write the $n$ integrals as $E[f(G_n)]$ for some random variable $G_n$ that satisfies $G_n\rightarrow 1/e$ almost surely. $\endgroup$
    – Michael
    Aug 3 at 14:12
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WLLN is sufficient but SLLN might be easier to think about.

You have that $\frac{1}{n} \sum_{i=1}^n \log X_i \to -1$ in probability. By the continuous mapping theorem for convergence in probability, this implies $$(X_1 \dots X_n)^{1/n} = \exp\left(\frac{1}{n} \sum_{i=1}^n \log X_i\right) \to \exp(-1) = 1/e$$ in probability, and likewise, $f((X_1 \dots X_n)^{1/n}) \to f(1/e)$ in probability. Now $f$ is bounded, and the dominated convergence theorem is valid for convergence in probability, so we do indeed get $\mathbb{E}[f((X_1 \dots X_n)^{1/n})] \to f(1/e)$.

If you use the strong law instead, then you don't have to think about the continuous mapping theorem, and you can use the more familiar version of the dominated convergence theorem.

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