# Thread: Help find counterexample: multiple limits

1. ## Help find counterexample: multiple limits

A true/false question from a quiz a while back. It's still bugging me because I'm told it's a false statement, but after three weeks I can't think of a counterexample. Maybe I just haven't quite made sense of it. A suitable counterexample or even a clarifying explanation of what this is saying would be of great help. Thanks in advance--here's the statement:

If $\displaystyle \int _0 ^{\infty} f_k(x) dx = \lim _{A\to \infty} \int _0 ^A f_k(x) dx$ exists for each $\displaystyle k \in \mathbb{N}$ and $\displaystyle f_k \to 0$ uniformly on $\displaystyle [0,\infty )$, then $\displaystyle \lim _{k\to \infty } \left(\lim _{A\to \infty } \int _0 ^A f_k(x) dx \right) = 0$.

2. I am very unsure so maybe somebody else can answer too, but I think it is about order of taking limits. I thought true at first too but I remember sometimes the order of limits can give different result. If the order of limits did not mater then you can change to take first the limit as k goes to infinity of the finite integral from 0 to A of f_k. This will be 0 for any A because f_k ---> 0 uniformly and it is ok to switch order of limit and integral when there is uniform convergence of sequence of functions. Then taking other limit as A goes to infinity will still be 0. You must find example that order of taking limits is different. I am sorry I can not make one I have been away from math many years that is why my name is memory loss LOL

3. ## Fading pulse example (updated)

Try this one.
Define $\displaystyle f_k(x) = \left\lbrace \begin{array}{cl} \frac{1}{k} & x \in [0,k] \\ 0 & \text{otherwise} \end{array} \right.$ then $\displaystyle f_k(x) \rightarrow 0$ uniformly on $\displaystyle [0,\infty)$ but for each k, $\displaystyle \int_{0}^{\infty}{f_k(x)} = \frac{1}{k} \times k = 1$ so $\displaystyle \lim_{k \rightarrow \infty}{\int_{0}^{\infty}{f_k(x)}} = 1$.

4. Originally Posted by huram2215
Try this one.
Define $\displaystyle f_k(x) = \left\lbrace \begin{array}{cl} \frac{1}{k} & x \in [0,k] \\ 0 & \text{otherwise} \end{array} \right.$ then $\displaystyle f_k(x) \rightarrow 0$ uniformly on $\displaystyle [0,\infty)$ but for each k, $\displaystyle \int_{0}^{\infty}{f_k(x)} = \frac{1}{k} \times k = 1$ so $\displaystyle \lim_{k \rightarrow \infty}{\int_{0}^{\infty}{f_k(x)}} = 1$.
Ooooh, that was slick. Much obliged!

So memory loss was right--it's about the order of the limiting processes mattering.