Let $f(x) = \mathrm{e}^{-x}\mathbb{1}_{[0, \infty)}$. I have a homework that asks me to approximate $f$ by a sequence of non-negative simple functions $f_n$ which is increasing and converges pointwise to $f$ and then compute the Lebesgue integral of $f$. I tried to follow the general construction, which can be found here but I appreciate other, perhaps much simpler constructions for this specific case.
Since $f$ only takes values in $[0,1]$ we only need to subdivide this interval into disjoint sets $A_{n,k}$. For each $n\in \mathbb{N}$ let $I_{n,k} = [\frac{k-1}{2^n}, \frac{k}{2^n})$ for $k = 2,\dots 2^n$ and $I_{n,1} = (0, \frac{1}{2^n})$. Define $A_{n,k} = f^{-1}(I_{n,k})$ and let $$f_n := \sum_{k=1}^n \frac{k-1}{2^n}\mathbb{1}_{A_{n,k}} + \mathbb{1}_{\{0\}}.$$ Then the sequence $(f_n)_{n \in \mathbb{N}}$ is increasing and converges pointwise to $f$. We have $$A_{n,1} = (-\infty, 0) \cup (-\ln(\frac{1}{2^n}), \infty)\\ A_{n,k} = (-\ln(\frac{k}{2^n}, -\ln(\frac{k-1}{2^n})),\quad k \neq 1, 2^n\\ A_{n,2^n} = (0, -\ln(\frac{2^n-1}{2^n})).$$
Using the definition of the Lebesgue Integral we obtain \begin{align*} \int_{\mathbb{R}}f\; \mathrm{d}\lambda^1 &= \lim\limits_{n \to \infty} \int_{\mathbb{R}} f_n\; \mathrm{d}\lambda^1\\ &= \lim\limits_{n \to \infty} \frac{1}{2^n}\sum_{k=1}^{2^n}(k-1)\lambda^1(A_{n,k})\\ &= \lim\limits_{n \to \infty} \frac{1}{2^n}\sum_{k=1}^{2^n-1}k\cdot \lambda^1(A_{n,k+1})\\ &= \lim\limits_{n \to \infty} \frac{1}{2^n}\sum_{k=1}^{2^n-1}k\cdot \ln(\frac{k+1}{k})\\ &= \lim\limits_{n \to \infty} \frac{1}{2^n}\ln\left(\prod_{k=1}^{2^n -1}\left(\frac{k+1}{k}\right)^k\right)\\ &= \ln\left(\lim\limits_{n \to \infty}\sqrt[2^n]{\prod_{k=1}^{2^n-1}\left(\frac{k+1}{k}\right)^k}\right)\\ &= \ln\left(\lim\limits_{n \to \infty}\sqrt[2^n]{\frac{\left(2^n\right)^{2^n-1}}{(2^n-1)!}}\right) \end{align*} where the last equality follows from continuity of the natural logarithm. Now if my construction is correct, the limit inside should be $\mathrm{e}$, so that we get $\int_{\mathbb{R}}f\; \mathrm{d}\lambda^1 = 1$ as desired. However I'm having trouble proving this. Does anyone have any idea?
The limit inside does in fact evaluate to e. First note that we can change all appearances of $2^n$ in the expression to $m$ and consider the limit as $m \to \infty$. If that exists, then so does the original limit (and their values are the same).
$$\lim_{n\to \infty} \sqrt[2^n]{\frac{(2^n)^{2^n-1}}{(2^n-1)!}} = \lim_{m\to \infty} \sqrt[m]{\frac{m^{m-1}}{(m-1)!}} = \lim_{m\to \infty} \sqrt[m]{\frac{m^{m}}{m!}} = \lim_{m\to \infty}\frac{m}{\sqrt[m]{m!}}\to e$$ as by Stirlings Approximation we have $m! \sim\sqrt{2\pi m} \left(\frac{m}{e}\right)^m$ for large $m$ and hence the last limit goes to $e$.