# Discussing a problem with Riemann integral.

As an introduction to Lebesgue integration, our professor gave us some problems of Riemann integration. One of these problems is the following function:

$$f_{n}(x) = n^2 x e^{-nx}.$$

He said the problem with this function is that:

$$\lim_{n \rightarrow \infty} f_{n}(x) = 0$$ for $$x \in [0,1]$$ And so the integration from $$0$$ to $$1$$ is also $$0.$$ But $$\int_{0}^{1} f_{n}(x)dx = 1.$$

My questions are:

1- Does the limit and the integral sign can always be interchanged in the case of Riemann integration? I do not think so, I think this is only true in case of uniformly continuous functions. Am I correct?

2- What is the reason for taking $$x \in [0,1],$$ is it for integrating reasons or is there a reason regarding the process of taking the limit?

3- Why did not we integrate over $$n$$ and not $$x$$?

4- How are we comparing integration over $$x$$ to taking the limit over $$n$$? Is not those are 2 very different things?

Could anyone help me in answering these questions that irritates my mind please?

EDIT:

Also, I calculated $$\int_{0}^{1} f_{n}(x)dx$$ but it was not 1 (I got $$\frac{-n}{e^n} - \frac{1}{e^n} + 1$$). am I correct? it is 1 after taking the limit as $$n \to \infty.$$

• Do you know what is Pointwise and Uniform convergence of a sequence of function $(f_n)$. If not, it will be a starting point. Commented Jun 16, 2020 at 15:01
• 1. NO. This is exactly the point of the example of your post. 2. No particular reason. The simpler a counterexample is, the better. 3. $n$ is an integer. How would you integrate wrt to an integer variable ? Here $f_n$ is a function of the real variable $x$ on a real interval, so we can integrate wrt to $x$. 4. The point of the example is to show you that you $\int_0^1 \lim f_n(x) dx\neq \lim_n\int_0^1 f_n(x) dx$. No more, no less. Commented Jun 16, 2020 at 15:02
• @hamam_Abdallah Yes I know them ... but I can not see how knowing them answers my concerns? could you explain more please?
– user778657
Commented Jun 16, 2020 at 15:11
• @GreginGre So in case of Lebesgue integration, the limit and the integral sign can always be interchanged?
– user778657
Commented Jun 16, 2020 at 15:15
• @GreginGre could you please explain more your answer No for my first question?
– user778657
Commented Jun 16, 2020 at 15:24

Integrating by parts, as you apparently have done, gives $$\displaystyle\int_0^1 f_n(x) \, dx = 1 - (n+1)e^{-n}$$ and, thus,

$$\lim_{n\to \infty} \int_0^1 f_n(x) \, dx = 1 \neq 0 = \int_0^1 \lim_{n\to \infty}f_n(x) \, dx$$

To show that $$\displaystyle\lim_{n \to \infty}f_n(x) =\lim_{n \to \infty}n^2x e^{-nx} = 0$$, note that $$f_n(0) = 0$$ and for $$0 < x \leqslant 1$$ we have

$$0 \tag{*}\leqslant n^2x e^{-nx} = \frac{n^2x}{e^{nx}} < \frac{n^2x}{\frac{n^3 x^3}{3!}} = \frac{6}{nx^2}$$

The inequality holds because $$\displaystyle e^{nx} = 1+ nx + \frac{(nx)^2}{2!} + \frac{(nx)^3}{3!} + \ldots> \frac{(nx)^3}{3!}$$.

The RHS of (*) converges to $$0$$ as $$n \to \infty$$ and, by the squeeze theorem, it follows that $$n^2x e^{-nx} \to 0$$.

• Is using L'hopital also correct?
– user778657
Commented Jun 17, 2020 at 1:47
• @Smart20: That will work as well.
– RRL
Commented Jun 17, 2020 at 2:08
• @Smart20: the interchange between limit and integral happens under very general conditions in case of Lebesgue integral. The Riemann integral is far more restrictive in this regard. But one can't interchange them always. Commented Jun 17, 2020 at 2:52
• Riemann integrals are defined for bounded functions and bounded intervals. If the Riemann integral exists, then so does the Lebesgue integral (and they coincide). So this example shows the interchange does not always work for the Lebesgue integral as well.
– RRL
Commented Jun 17, 2020 at 2:53
• If $(f_n)$ is a sequence of Lebesgue integrable functions on some $E \subset \mathbb{R}$ and $f_n \to f$, then under certain conditions we can be sure that $\int_E f_n \to \int_Ef$. The monotone convergence theorem and dominated convergence theorems give sufficient conditions for the interchange of the limit and the Lebesgue integral.
– RRL
Commented Jun 17, 2020 at 2:56