# Definitions of Lebesgue integral

1. Jan 10, 2015

### DavideGenoa

Dear friends,I know the definition, from A.N. Kolmogorov and S.V. Fomin's Элементы теории функций и функционального анализа, of Lebesgue integral of measurable function $f:X\to \mathbb{C}$ on $X,\mu(X)<\infty$ as the limit

$\int_X fd\mu:=\lim_{n\to\infty}\int_Xf_nd\mu=\lim_{n\to\infty}\sum_{k=1}^\infty y_{n,k}\mu(A_{n,k})$​

where $\{f_n\}$ is a sequence of simple, i.e. taking countably many (not necessarily finitely) values $y_{n,k}$ for $k=1,2,\ldots$, functions $f_n:X\to\mathbb{C}$ uniformly converging to $f$, and $\{y_{n,k}\}=f_n(A_{n,k})$ where $\forall i\ne j\quad A_{n,i}\cap A_{n,j}=\emptyset$. If $\mu$ is not finite but is $\sigma$-finite with $X=\bigsqcup_{j=1}^\infty X_j$ then $\int_{X}fd\mu:=\sum_{j=1}^\infty\int_{X_j}fd\mu$, provided that $\sum_{j=1}^\infty\int_{X_j}|f|d\mu<\infty$.

I read other authors defining the Lebesgue integral for non-negative functions $g:X\to[0,+\infty]$, where $\mu(X)\le\infty$, by using simple non-negative functions $s_i$ taking only finitely many values on measurable sets $A_i$ in the following way:

$\int_X gd\mu:=\sup\{\sum_i s_i\mu(A_i):\forall x\in X\quad s_i(x)\le g(x)\}$
and in general in the following way, provided that $\int_X|f|d\mu<\infty$:

$\int_Xfd\mu:=\int_X \text{Re}f_+d\mu-\int_X \text{Re}f_-d\mu+i(\int_X \text{Im}f_+d\mu-\int_X \text{Im}f_-d\mu)$
where subscript $\pm$ is used to define $g_+(x):=\max\{g(x),0\}$ and $g_-(x):=-\min\{g(x),0\}$ for any measurable function $g$.

I have not been able to find more about it, since the texts that are available to me all use Kolmogorov-Fomin's definition, but I am convinced that such definitions are equivalent. From what I know -I have finished Kolmogorov-Fomin's Элементы теории функций и функционального анализа, which significantly overlaps with the English language translation Introductory Real Analysis by the same authors- I would say that it would be enough to prove the equivalence for non-negative functions $g:X\to[0,+\infty)$.
If they are equivalent, how can it be proved?
I uncountably thank you and wish you a happy 2015!

2. Jan 10, 2015

### mathwonk

can you prove that a function with countably many values has an integral which is a limit of functions with finitely many values? (that sounds essentially like the definition of a limit of a sequence of real numbers.)

3. Jan 10, 2015

### DavideGenoa

Thank you!
Well, the set of measurable functions taking finitely many values on sets of finite measure is dense in $L^1(X)$ (where I use Kolmogorov-Fomin's definition of Lebesgue integrability for $L^1$)...
I had thought that could be used, but I have not been able to use it to prove the desired equivalence...

4. Jan 10, 2015

### Stephen Tashi

5. Jan 10, 2015

### mathwonk

basically the space of integrable functions is defined as the closure of the countable valued ones, and the integral is defined as the limit of those integrals. saying the countable valued ones are limits of finite valued ones seems to say the finite valued ones have the same closure, so you would get the same space of integrable functions. then the limit theorems would seem to say the integral of a function in the closure is again the limit of the integrals of the approximating finite valued functions.

there seem to be two ways to define the lebesgue integral, depending on how much measure theory you want to use. one way, the favorite of my analyst friends, is to define, (say in the real numbers), measurable sets, and the measure of a measurable set, using inner and outer measures. Then one defines measurable functions, analogous to continuous ones, by saying they pull back measurable sets to measurable sets.

finally one defines the integral of a measurable function as a limit of integrals of simple, or countable valued, measurable functions with finite integrals. There are then various theorems describing what families of functions are dense in the space of integrable functions (measurable functions whose integrals are finite). Crucially, there are theorems that say for a pointwise convergent sequence, when the limit of the integrals is the integral of the limit.

If one wants to avoid all the measure theory, one can just choose a nice subfamily of more easily integrable functions (known to be dense), and define the space of integrable functions as the limits of these functions, and integrals to be limits of these integrals. However in this approach, not having defined the integral in an intrinsic way independent of the approximating sequence, one must prove that all approximating sequences of the desired type have the same limiting integral.

Dense subfamilies of functions used for this approach may even be smooth functions, and I believe one may choose approximation initially from below, then above, or a combination of both. In this way one only needs to know the definition of Riemann integration, since this suffices for the dense subfamily, and then just says that for more comp-licated functions the integral is define by choosing an approximating sequence and taking the limit of their Riemann integrals.

Basically two definitions are equivalent if they give the same answer for all integrable functions, and also define the same space of integrable functions. Or if they give different spaces but the same answer on their overlap, you can say they are equivalent where the makes sense. E.g. Riemann and Lebesgue integration are equivalent on the space of continuous, or Riemann integrable functions.

It is often pointed out as a limitation of the Riemann integral that the pointwise limit of Riemann integrable functions may not be Riemann integrable. However, it is still possible to use such limits to define the integral of the limit function. I.e. one does not care if the limit is Riemann integrable or not. What matters is that for all pointwise convergent sequences to the given limit function, whose Riemann integrals converge, all the sequences of Riemann integrals have the same limit. The integral of the limit function can then be defined as the limit of those approximating integrals. This is essentially the second, approximation approach, mentioned above to Lebesgue integration.

(technical remark/woffle: it may be useful/necessary to restrict some statement by the phrase "almost everywhere". I am not an analyst and have not taught this material since 1970, or thought it fully through in the approximation form.)

Summary: Given a suitable function, one can define its Lebesgue integral using measure theory, and then prove that its integral is the limit of the integrals of any approximating sequence of Riemann integrable functions which converge to it pointwise; or one can simply choose a sequence of Riemann integrable functions which converge to it pointwise (if such exist), and define its integral as the limit of those Riemann integrals. Then of course one must prove that if two such sequences converge pointwise to the same function, then their Riemann integrals have the same limit.

This is a simplified version of the correct statements due to my ignorance, but I hope it gives some useful information.

As to the original question, if two definitions give the same answer on a dense set, and both satisfy the " limit of the integrals is the integral of the limit" theorems, then they give the same answer everywhere they both make sense.

Last edited: Jan 10, 2015
6. Jan 11, 2015

### DavideGenoa

Wow: very interesting background! Thank you!!!

I am not sure I understand what metric is intended... I $\infty$-ly thank you!

7. Jan 11, 2015

### mathwonk

the metric is called the L1 metric. i.e. fn converges to f if the integrals of the absolute value of the differences, |f-fn|, converges to zero.

In fact one definition of the space of integrable functions is just to take the abstract metric completion of some nice family of functions, such as simple ones. But then one has to prove that each such new point of the abstract completion isn actually a function, by proving that an L1 cauchy sequence of simple functions converges also pointwise almost everywhere to a function.

8. Jan 11, 2015

### DavideGenoa

Thank you for the remark!

9. Jan 13, 2015

### mathwonk

Last edited by a moderator: May 7, 2017
10. Jan 26, 2015

### lavinia

While the general definition of a measurable function is that inverse images of measurable sets are measurable, for lebesque measure this is the same as saying that inverse images of open sets are measurable. This definition allows a simple and intuitive definition of the Lebesque integral.

For simplicity assume for the moment that the function is nonnegative. (This is not really a significant restriction since the case of positive and negative and of complex valued functions is defined in terms of integrals of positive functions.)

Divide the positive y-axis up into half open intervals with the lower end point included in the interval.

The inverse image of each half open interval is measurable. Take the function whose value on the inverse image of each half open interval is the value of the measurable function,f, at each lower endpoint. This defines a simple function that is less that or equal to f at all points.

If one chooses the intervals to be of length less than ε, then the simple function will be everywhere within ε of f. Thus a sequence of ε's that converges to zero will produce a sequence of simple functions that converge uniformly to f. This ,I think, is where the uniform convergence condition comes from.

Integration theory is then defined by letting the integral of f be the sup of the integrals of simple functions below it. I found this definition a little deceptive until I read about constructing this sequence of simple functions by dividing up the y-axis into half closed intervals.

In a way, the idea of Lebesque integration is the opposite of Riemann integration in the sense that Riemann integration divides the x axis into disjoint intervals while Lebesque integration divides the y axis into disjoint intervals.

This definition extends naturally to real valued and then complex valued functions. So the two definitions are really identical.

Last edited: Feb 7, 2015
11. Jan 26, 2015

### DavideGenoa

Thank you so much for the answer, Lavinia!

12. Feb 5, 2015

### mathwonk

I like the description Lavinia gave very much. I would mention the contrast with the ones I described. In Lavinia's conceptually simple, and very clear definition, in practice it seems one must know the integral of all the approximating simple functions in order to compute or define the integral of the given function. This means one must know how to define measurable sets and to compute their measure. This involves coverings by intervals, and computing and equating inner and outer measures.

The alternate approach may be less intuitive, but may have the advantage that one knows from Riemann integration how to compute the integrals of the approximating functions. On the other hand it may be abstractly less clear how to actually choose those approximating functions. In practice however it seems to me one may commonly be given the function one wants to integrate as such a limit.

Here is a simple example, the characteristic function of the rational points in [0,1]. This is already a simple function so its integral is defined as the measure of the rationals in [0,1] times 1 plus the measure of the irrationals times zero, i.e. zero, once one knows the measure of a countable set is zero, and additivity of measure.

In the Riemann approximation approach one enumerates the rationals, and approximates our function by the corresponding characteristic functions of finite subsets of rationals, all of which integrals are zero, hence the limit is zero.

I guess I admit I do like the measure theory point of view here as somehow simpler, but I think it takes more work to compute.