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My professor asked me how to intuitively understand Lebesgue's dominated convergence theorem and what's the effect of the integrable dominated function. More specifically, when we are given a Lebesgue integrable function, why it suffices to consider other things on a subset with finite measure?

I think the professor means the following theorem:

If $f$ is Lebesgue integrate in $X$, for every $\epsilon>0$, there exists measurable $E$ with $\mu(E)<\infty$ s.t. $\int_{X\setminus E}|f|<\epsilon$.

Hence I proved the statement using monotone convergence of $|f|\chi_{B(x_0,n)}$, but it seems my professor was not satisfied. She wanted me to explain what's the essential difference between a Lebesgue integrable function and a Riemann integrable function, and what makes a Lebesgue integrable function not Riemann integrable. She said answering the second question will help me answer the original question.

I think from the definition of Riemann integral and the theorem, the Lebesgue integrable function is Riemann integrable if and only if it's continuous, i.e. Riemann integrability requires the function value to not have a large oscillation when $x$ changes slightly. Is it a correct explanation? How about Lebesgue integral? And how is it related to the first question?

I know this question may be a little subjective and may not have the so-called "correct" answer as other questions on the website. But I am still looking forward to your comments.

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  • $\begingroup$ So many views, so little activity. A dead, but seemingly good post. $\endgroup$ Dec 31, 2016 at 21:47

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Edit 1: I originally wrote that Riemann integrable functions are those with $countable$ sets of discontinuity at most. This is wrong, as @Ted Shifrin pointed out. Take the indicator function of the Cantor set: it has uncountably many discontinuity points, but they form a set of measure zero - the Cantor set itself.

So, we have two questions from your professor:

  1. What's the essential difference between a Lebesgue integrable function and a Riemann integrable function?
  2. What makes a Lebesgue integrable function not Riemann integrable?

Question 1 was essentially answered by you:

Riemann integrability requires the function value to not have a large oscillation when $x$ changes slightly.

This is true. We ask, for Riemann integrable functions, that they're discontinuous in sets with Lebesgue measure zero in $\mathbb{R}^{n}$, and thus "negligible". We could also add the fact that all Riemann integrable functions are bounded,whereas Lebesgue integrable functions need not be so. But that's essentially their difference, I believe.

As for question 2, you have to see why the function we have in our hands is not Riemann integrable in the first place. Is it (a) just not bounded? Or is (b) the set of its discontinuity points of positive Lebesgue measure? If (a), it isn't Riemann integrable because it just doesn't satisfy the definition of a Riemann integrable function. Now, if (b), we once again go back to what you said yourself: it oscillates too much in a given neighborhood of $x$.

As your professor said, answering the second question gives us a clearer answer to the first: the essential difference between Lebesgue and Riemann integrable functions is that, in general, the former is not as well-behaved as the latter.

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  • $\begingroup$ Are you sure about the countable set of discontinuities? $\endgroup$ Feb 25 at 16:26
  • $\begingroup$ Gonna have to rewrite that one, my bad. Forgot about the Cantor set $\endgroup$ Feb 25 at 16:58

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