# Using the law of large numbers to calculate an integral

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!

• 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. Aug 3 at 14:12

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 $$\mathbb{E}[f((X_1 \dots X_n)^{1/n})] \to f(1/e)$$.