# I Limitations of the Lebesgue Integral

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1. Jun 2, 2017

### The_eToThe2iPi

So I'm studying a course on measure theory and we've learnt that the Lebesgue integral of a real function is (loosely) defined as the total area over the x-axis minus the total area under the x-axis. This seems to me to be limited because these areas can both be infinite but their difference may be finite (such as sin(x)/x integrated from 0 to +inf) and, for me, this is a major failing for a definition of the integral. I was wondering whether there were any extensions to the Lebesgue integral that can handle these types of functions?

2. Jun 2, 2017

### zwierz

A function $f(x)$ is Lebesgue integrable iff $|f(x)|$ is Lebesgue integrable

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6. Jun 2, 2017

### Staff: Mentor

Don't you already have that for ordinary Riemann integrals $\int_0^\infty x\,dx \; , \;\int_1^\infty x\,dx$ and their difference? So what's new?

7. Jun 2, 2017

### The_eToThe2iPi

Ahh, that's interesting. I didn't think about that. But I still feel that there is a difference between the two instances. In your case we are evaluating two separate Riemann integrals, finding they are infinite, and then asking what their difference is like asking what is $(a+\infty) -\infty$. The Lebesgue integral is finding a single integral by separating the function into two parts and, as a result of this, means that certain functions in certain domains are not Lebesgue integrable even though they intuitively should be integrable. Obviously the question of what the "difference" between two infinite values can be stated in terms of Riemann integrals but it is not per say a failing of Riemann integration since the definition does not rely on this.
Also I apologise for the lack of rigour but I'm simply trying to further understand the concepts involved in integration and, whilst I appreciate the necessity for formalism, I'm still kinda a noob at this and trying to formulate every sentence in full rigour is both difficult and not very revealing...

8. Jun 2, 2017

### WWGD

I don't know if this helps, but you can also see the Lebesgue integral as partitioning the range , instead of the domain and finding the sum ( as a limit) $x *m(f^{-1} (x))$ , i.e., the sum of values in the range times the measure of their preimages. It is often put as, if you had 1 nickel ( 5c ) 2 dimes ( 10 c each), 3 quarters (25 c each) , the Riemann sum would be 5 + 10 +10 +25+ 25+ 25, while Lebesgue would do: 5(1)+ 10(2) +25(3) , with 1,2,3, being the (discrete version) of the measure of the preimage of a value.

9. Jun 2, 2017

### Staff: Mentor

Well, the Insight article I quoted is a good start and explains what @WWGD just said. In general measure theory can be very abstract (depends on the teacher and/or textbook). Basically the $"dx"$ becomes the subject of consideration and generalization rather than geometric objects. You might remember, that already $"dx"$ had been a generalization from something that started with length times width.

10. Jun 2, 2017

### The_eToThe2iPi

Wow, just imagine if students had to buy all their own textbooks! £152!!! Fortunately I think the university has a copy of that one so I won't have to sell my kidney :)
The page on guage integration mentions that this integral can be extended more general domains and I was wondering whether whether these extensions allow integration over domains as general as the Lebesgue Integral. Also are there not a similar examples of more general functions that the Lebesgue integral cannot handle or is it only for real functions that we encounter this problem.

11. Jun 2, 2017

### The_eToThe2iPi

To be honest, I feel like these are the sort of questions that require me to read a book (and I'll probably have even more afterwards). Thank you all so much for answering my questions and pointing me in the right direction.

12. Jun 2, 2017

### WWGD

Anyway, the standard example, in my experience of a non-Lebesgue-Integrable function is $f(x)= \frac {1}{x}$

13. Jun 2, 2017

### The_eToThe2iPi

What domain are you integrating $f(x)=\frac{1}{x}$ over? Does an improper Riemann integral exist for $f(x)=\frac{1}{x}$ over that domain? The reason I used $sin(x)/x$ was because it is an alternating function and so needs to be broken down into it's positive and negative parts in order to evaluate the Lebesgue integral and as a result this integral cannot be evaluated (in the Lebesgue sense) over the interval $[0,\infty)$. I cannot see this same problem with $f(x)=\frac{1}{x}$...

14. Jun 2, 2017

### WWGD

$(0,1]$ is an example. EDIT Compare, e.g., with $\int_{1/n}^1 \frac{1}{x} dx$ as lower bounds for the integral over $(0,1]$.

15. Jun 2, 2017

### The_eToThe2iPi

But doesn't this also not have a Riemann integral over that same domain since $\int^{1}_{\delta}\frac{1}{x}dx=[ln(x)]^{1}_{\delta}=ln(\frac{1}{\delta})$ but $ln(\frac{1}{\delta}) \to \infty$ as $\delta \to 0$?

Last edited: Jun 2, 2017
16. Jun 2, 2017

### WWGD

Yes, but that is the point; if the Riemann integral exists, then so does the Lebesgue, but in many cases some functions that are not R-integrable are L-integrable, but this one is neither.