1

How to use Lebesgue's Dominated Convergence Theorem and Monotone Convergence Theorem in Lebesgue Integration Theory to show equality ibetween Riemann Integration and Lebesgue integration?.

  • For each fixed x, the integrand converges to $x^n e^{-x}$ and x is eventually in the domain of integration. Thus you can use theorems from Lebesgue integration to conclude that your limit is the same as $\int_0^\infty x^n e^{-x} dx$. – Ian Dec 01 '17 at 16:20
  • For a continuous function on a bounded interval $[0,k]$ the classical and Riemann and Lebesgue integrals are equal. – DanielWainfleet Dec 02 '17 at 00:29

2 Answers2

3

HINTS:

For all $0\le x\le k$

$$\left(1-\frac xk\right)^k\le e^{-x}$$

Note that for the improper Riemann integral

$$\lim_{L\to \infty}\int_0^L x^ne^{-x}\,dx=n!$$

And finally, we have

$$\lim_{k\to \infty}\int_0^k x^n\left(1-\frac xk\right)^k\,dx=\lim_{k\to \infty}\int_0^\infty x^n\left(1-\frac xk\right)^k\,\xi_{[0,k]}(x)\,dx$$

where $\xi_{[0,k]}(x)$ is the indicator function and is bounded by $1$.

Mark Viola
  • 179,405
2

I think you want to know why applying something on the Lebesgue integral would let you conclude the same for the Riemann integral.

Let us start like this you want to calculate: \begin{align} \lim_{k\to \infty}\int_0^\infty f_k(x) dx \end{align} Where $f_k(x) = x^n(1-\frac{x}{k})^k \mathbf{1}_{[0,k]}$. This function is Riemann integrable on $[0,k]$ for all $k$. Furthermore we know that the Riemann integral of $f_k(x)$ on $[0,k]$ equals the Lebesgue integral (you should know the theorem that says so! See for example this post.). Furhter note that we can integrate from $0$ to infinity since $f_k(x)$ is zero for $x>k$. I will denote $\int dx$ for Riemann integral and $\int d\mu(x)$ for the Lebesgue integral. So we have: \begin{align} \int^\infty_0 f_k(x) dx=\int^\infty_0 f_k(x) d\mu(x) \end{align} So this means: \begin{align} \lim_{k\to\infty} \int^\infty_0 f_k(x) dx = \lim_{k\to\infty}\int^\infty_0 f_k(x) d\mu(x) \end{align} It is beautiful, because now you can use what @Mark Viola have written in his answer to the integral on the RHS so you can conclude that for your original Riemann integral.

Shashi
  • 8,738
  • why do you say $f_k(x)$ is zero for $x>k$? I think $f_k(x)$ is zero for $x<k$, is it @Shashi? –  Dec 01 '17 at 17:21
  • @Imhejhey no $f_k(x)$ is zero for $x>k$ because $f_k(x)$ is $x^n(1-\frac{x}{k})^k$ multiplied by the indicator function $\mathbf{1}_{[0,k]}$. And the indicator function that I have used is $1$ for $x\in[0,k]$ and zero otherwise. – Shashi Dec 01 '17 at 17:24
  • oh ya, my mistake –  Dec 01 '17 at 17:26