# Moving a limit inside an integral

1. Oct 7, 2007

### strangequark

A grad student mentioned the other day that you cannot move a limit inside of an integral without meeting certain conditions, unfortunately, he didnt say what those condition were... I was under the impression that this was unrestricted (and the particular theorem we were looking at worked fine if we did this)...

Can anyone name these conditions for me?

2. Oct 7, 2007

### arildno

A typical case is the following:
Let fe(x)=0, |x|>=e>0, f(x)=1/(2e), |x|<e.

Thus, we have:
$$I_{e}=\int_{-\infty}^{\infty}f_{e}(x)dx=1, \lim_{e\to{0}}I_{e}=1$$

However, the following limit doesn't EXIST, $$\lim_{e\to{0}}f_{e}(x)$$
Thus, it is inadmissible to interchange the limiting operations of integration and sequence limiting.

3. Oct 7, 2007

### CompuChip

From my notes from an Analysis course:

Let $V \subset \mathbb{R}^n, a, b \in \mathbb{R}, a < b$. Further, assume that the function $f: V \times [a, b] \to \mathbb{R}$ is continuous on $V \times [a, b] \subset \mathbb{R}^{n + 1}$. Then the function $I: V \to \mathbb{R}$, defined by
$$I(x) := \int_a^b f_x(t) \, \mathrm{d}t = \int_a^b f(x, t) \, \mathrm{d}t, \qquad f_x: t \mapsto f(x, t)$$
is continuous.

This means that for each $\xi \in V$,
$$\lim_{x \to \xi} \int_a^b f(x, t) \, \mathrm{d}t = \lim_{x \to \xi} I(x) = I(\xi) = \int_a^b f(\xi, t) \, \mathrm{d}t = \int_a^b \left( \lim_{x \to \xi} f(x, t) \right) \, \mathrm{d}t$$,
using the continuity of $x \mapsto f(x, t)$ in the last step.

The proof is not hard, but not easy to type out since it relies on all sorts of definitions and earlier results (e.g. on interchange of limits and results on Riemann integration).

4. Mar 19, 2009

### zrezki

CompuChip, can you please elaborate on the arguments behind the proof (a sketch would be enough). Thanks!

5. Mar 19, 2009

### GSpeight

The monotone convergence theorem and dominated convergence theorem form measure theory can often be used to pass the limit inside an integral, rephrasing convergence in terms of convergence of sequences when necessary.

6. Mar 19, 2009

### zrezki

Sorry, there is no sequence or series in the equatiion below:

$\underset{x \rightarrow \infty}\lim \int_{0}^{\infty} f(x,y)dy = \int_{0}^{\infty} \underset{x \rightarrow \infty}\lim \Bigl[ f(x,y) dy$

assuming f is continuous everywhere in $[a,\infty[ \times [0,\infty$ and the limit inside the integral on the above equation exists and is finite. According to Compuchip the above statement holds and my question what are the mathematical arguments.

Thanks,

zrezki

7. Mar 19, 2009

### lurflurf

You we under a bad impression.
The integral is not important.
The issue is interchanging two limits.
Or more precisely when interchanged limits exist and are equal.
say U and V are limits and f a function for what U,V and f is
UVf=VUf
It is a hard question to answer in general.
In simple cases a sufficient condition is used.
Limits can be interchanged when uniform or close to it.
That is to say the inside limit must not change rapidly as the outer limit is approched.

8. Mar 19, 2009

### zrezki

Still heuristic and I need more rigourous arguments. This is a well-formed problem to which should exist a clear answer.

Thanks

9. Mar 19, 2009

### lurflurf

^This problem is not well formed and no clear answer exist. You did not specify what rigorous means. In particular the type of integration has not been specified nor the conditions on the function to be integrated, nor the nature of the limit being taken.

Here is an example of a well-formed problem to which a clear answer exists.

let I indicate (proper) Riemann integration with respect to x with limits a and b
let L indicate a limit operation with respect to t

let the function f(x,t) be continuous in x for the range a<=x<=b for all t in the approach,
and if f(x,t) approaches a limiting function F(x) uniformly in x for this range, then:
ILf and LIf exist and are equal

notice that one can extent this if desired and even then all cases are not settled

we will have (by previous theorem F is continuous hence integrable
ILf=IF exist
by uniform convergence for epsilon>0 there exist tepsilon
|F(x)-f(x,t)|<epsilon for t beyond tepsilon, a<=x<=b
so
|IF(x)-If(x,t)|=|I[F(x)-f(x,t)]|<=I|F(x)-f(x,t)|<=|b-a|epsilon
hence
LIf exist and
ILf=LIf

10. Mar 19, 2009

### zrezki

All the ifs you mentionned were contained in my previous reply. The only thing I forgot is that the integral is Riemann one. Thar's what I meant by well formed problem. Following your reasonning in the first part, all we need to move the limit inside the integral is unform convergence of f with respect to x. Is that true ?

11. Mar 19, 2009

### jostpuur

The comment from GSpeight seems to be the only one that answers the original question properly. The Lebesgue's dominated convergence theorem is the standard tool for changing the order of integration and limit.

The value of the Riemann integral is the same as the value of the Lebesgue integral, so of course the dominated convergence can be used for Riemann integrals!

12. Mar 19, 2009

### zrezki

I couldn't see how may I gonna apply the dominated convergence theorem to check if this is true:

$\underset{x \rightarrow \infty}\lim \int_{0}^{\infty} f(x,y)dy = \int_{0}^{\infty} \underset{x \rightarrow \infty}\lim \Bigl[ f(x,y) \Bigl}dy$

assuming f is continuous everywhere in $[a,\infty[ \times [0,\infty$ and the limit inside the integral on the above equation exists and is finite.

Thanks

13. Mar 19, 2009

### jostpuur

This hypothesis is not true. For example set

$$f(x,y) = \left\{\begin{array}{ll} 0,\quad &y<x\\ \sin(y-x),\quad & 0\leq y-x\leq \pi\\ 0,\quad &x+\pi < y\\ \end{array}\right.$$

$$\int\limits_0^{\infty} f(x,y) dy = \int\limits_{x}^{x+\pi} \sin(y-x)dy = 2\quad\quad\forall\; x\in [0,\infty[$$

$$\lim_{x\to\infty} f(x,y) = 0,\quad\quad \forall\; y\in [0,\infty[$$

The dollar signs \$ don't work over here. Instead use [ tex] and [ /tex].

Last edited: Mar 19, 2009
14. Mar 19, 2009