Let $(f_n)_{n\in\mathbb{N}} \rightarrow f$ on $[0,\infty)$. True or false: $\lim_{n\to\infty}\int_0^{\infty}f_n(x) \ dx = \int_0^{\infty}f(x) \ dx.$

Question

The Assignment:

Let $(f_n)_{n\in\mathbb{N}}$ converge uniformly to $f$ on $[0,\infty)$ and let the improper integrals of $f_n$ and $f$ exist on $[0,\infty)$. True or false: $$\lim_{n\to\infty}\int_0^{\infty}f_n(x) \ dx = \int_0^{\infty}f(x) \ dx$$ Prove it.

Since I know that under certain circumstances (if $f_n$ is integrable $\forall n≥1$) on the closed interval $[a,b]$: $$\lim_{n\to\infty} \int_a^bf_n(x) \ dx = \int_a^bf(x) \ dx$$ I have the tendency to say that the statement above is false, since these "safe" conditions (compact interval) are substituted by others.

Any help would be appreciated

Answer 1

Hint:

Fill a rubber square aquarium with water until it's completely filled. Now, take the right edge of the aquarium and stretch it to the right. The amount of water never changes, even though the height of the water decreases. What happens when you keep stretching it farther and farther to the right, without restriction?

Edit:

Since I answered your question with a riddle, I'll add a useful tip that might seem more math-like:

Whenever you encounter the question

Does condition XYZ guarantee $\displaystyle \lim_{n\to\infty} \int f_n = \int \lim_{n\to\infty} f_n$ ?

it is helpful to make the following observation:

If it is true in the special case that $f_n \to 0$, then it is true anyway.

This observation helps you in that you only really need to prove this case, or to find a counterexample of this sort.

Proof: Suppose it is true in the special case $f_n \to 0$. Consider another case, where $f_n \to f \ne 0$. Define the sequence $g_n = f_n - f$. Then $g_n \to 0$, and condition XYZ probably holds for $g_n$ as well as it does for $f_n$, so you can get $\lim \int (f_n - f) = 0$, which is exactly what you want.

Answer 2

Consider the flattening mountain of constant integral

$\begin{array}{ccccc} f_n & : & R^+ & \to & R^+\\  & & x & \mapsto &\frac{x}{n^2}-1+\frac{1}{n} \; \text{when} \; n^2-n<x\leq n^2\\ && x & \mapsto & \frac{-x}{n^2}+1+\frac{1}{n} \; \text{when} \; n^2<x<n^2 +n\\ && x & \mapsto & 0\text{when} \; n^2-n>x \; \text{or} \; x>n^2 +n \\ \end{array}$

Each $f_n$ has integral $1$ but the sequence converges uniformly to $0$.

Answer 3

Consider $f_n:=\frac{1}{n}\chi_{[n,2n]}$ where $\chi$ denotes the indicator function. Obviously $||f_n||_\infty\to 0 $, but $\int_0^\infty f_n=1~\forall n\in\mathbb{N}$.