Can I apply Monotone Convergence to $\int_{0}^\infty e^{-x} dx = \lim_{n \to \infty} \int_{0}^{n} (1 - x/n)^n dx = 1$ Can I apply Monotone Convergence to $\int_{0}^\infty e^{-x} dx = \lim_{n \to \infty} \int_{0}^{n} (1 - x/n)^n dx = 1?$
So the sequence $f_n= (1 - x/n)^n $ is nonnegative and increasing and converges point wise to $e^{-x}$. Everything seems to be setting up for the theorem.
But I have two questions


*

*Since the limit of integration $\int_{0}^n$, is it allowed to apply the theorem when the index are also increasing? The theorem is stated without any set in the integral

*Suppose (1) is resolved and the theorem can be applied. Why am I allowed to evaluate $\int_{0}^\infty e^{-x} dx$ as a Riemann integral? How do I know its Lesbegue $(L)\int_{0}^\infty e^{-x} dx$ coincides with its $R\int_{0}^\infty e^{-x} dx$? Simply because the limits in the Monotone Convergence theorem exists? And therefore the Lesbegue integral exists and coincides with Rieman Integral?
 A: Using Lebesgue integral with respect to a measure $\mu$ you can write
$$\int_{(0,\infty)}e^{-x}d\mu=\int \lim_{n \to \infty} 1_{[1/n,n]}(x)(1 - x/n)^n d\mu $$
Since $1_{[1/n,n]}(x)(1- x/n)^n\geq 0$ and it's increasing with respect to $n$ (check the derivatives) you can apply Beppo-Levi theorem yielding 
$$\int_{(0,\infty)}e^{-x}d\mu= \int \lim_{n \to \infty} 1_{[1/n,n]}(x)(1 - x/n)^n d\mu =\lim_{n\rightarrow \infty}\int 1_{[1/n,n]}(x)(1 - x/n)^n d\mu$$
Since the function $(1 - x/n)^n$ is continuous  and bounded in $[1/n,n]$ we know that it's Riemann itegrable in that interval. We know that if the Riemann integral exists then the Lebesgue integral exists and they coincide. 
Hence we can write the latter as
$$\lim_{n\rightarrow \infty} (\mathfrak R)\int_{1/n}^n (1 - x/n)^ndx$$
If this limit exists then the improper Riemann integral must coincide with the Lebesgue integral, namely$(\mathfrak R)\int_0^{\infty}e^{-x}dx=\int_{(0,\infty)} e^{-x}d\mu$.
[I am using Theorems 9.6, 11.8 and corollary 11.9 from "Measures, Integrals and Martingales" by Robert Schilling. ]
