Is it true that lower and upper Lebesgue integrals are equal for every Lebesgue integrable function?

If you look here you will see that:

One nice feature of measurable functions is that the lower and upper Lebesgue integrals can match, if one also assumes some boundedness.

Exercise 11 Let $f: {\bf R}^d \rightarrow [0,+\infty]$ be measurable, bounded, and vanishing outside of a set of finite measure. Show that the lower and upper Lebesgue integrals of $f$ agree. (Hint: use Exercise 4.) There is a converse to this statement, but we will defer it to later notes. What happens if $f$ is allowed to be unbounded, or is not supported inside a set of finite measure?

So... what happens if $f$ is unbounded? Are upper and lower integrals equal? One person in the comments suggested it's true for unbounded functions as well (but he didn't provide a proof).

Lebesgue defines area - then it's reasonable to expect that approximation of the area 'from above' and 'from below' should yield the same value (it's equivalent to saying that upper and lower Lebesgue integrals are equal - it would be nice if it was true).

Based on the comments to my question, I'd say it depends on the definition of simple function - whether we want it to take finite valus, or countably many values. But, according on the definition in prof. Tao's article, what would be the correct answer to question raised in Exercise 11, quoted above?

Apparently there are examples of Riemann integrable functions, whose upper Lebesgue integral doesn't equal (lower) Lebesgue integrals. But it's true that if both Riemann and Lebesgue integrals exist, they are equal.

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    $\begingroup$ Using the definitions in the book, I don't see how upper Lebesgue integrals could exist for an unbounded function. What simple function could dominate an unbounded function? What would be the finite set of values? $\endgroup$ – T.J. Gaffney Oct 12 '15 at 17:03
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    $\begingroup$ Rather, the upper integral would be infinite, because that's the infinimum of the empty set of simple functions which dominate. But not all unbounded functions have infinite Lesbesgue integral. $\endgroup$ – T.J. Gaffney Oct 12 '15 at 17:05
  • $\begingroup$ @Gaffney you're saying 'what simple function could dominate an unbounded function'? Well, can't we just define a simple function which is greater or equal than $f$ for all $x$? $\endgroup$ – user4205580 Oct 12 '15 at 20:02
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    $\begingroup$ If $f$ is unbounded, then all the "dominant" simple functions would be infinite on a set of non-zero measure. All the "dominant" integrals would then be infinite. $\endgroup$ – bartgol Oct 12 '15 at 20:59
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    $\begingroup$ If you define a simple function as a function which takes on only a countable number of values, then you can find an upper bound with finite integral, but that would be a definition of simple that the book doesn't use. $\endgroup$ – user24142 Oct 13 '15 at 19:26

According to the definition 1 simple functions are measurable functions that take only afinite number of values.

take the function $f:x\mapsto\frac{1}{x^2}\mathbb{1}_{[1;+\infty[}$ (from R to R) and take $g$ a simple function such that $g\geq f $. We call $\alpha$ the smallest non zero value of $g$ (which exists because $g$ only take a finite number of values), since $g\geq f$ we know that for every $x\geq 1$ we have $g(x)>0$ and so $g(x)\geq \alpha$. Thus $\int g \mathrm d\lambda\geq \alpha \cdot m([1;+\infty[)=+\infty$. So the upper Lebesgue integral of $f$ is infinite whereas the lower lebesgue integral of $f$ is equal to the Riemann integral of $f$, which is finite.

The same thing can be done with $f : x \mapsto \frac{1}{\sqrt x}\mathbb{1}_{]0;1]} $.

This two counterexample cease to work if we allow a simple function to take a countable number of values.

EDIT (in response to the edit of the question) :

First, what you say is not correct, the functions i've used in my example are not riemann integrable strictly speaking. If you use Riemann integration it'll be "improper" integrals. Usually (but it might varie a little between different autors) Riemann integral deal only with bounded functions with compact support, for which upper and lower lebesgue integral are equal (and for which riemann and lesbesgue integral are equal)

Second and last, in the article you linked Tao define the unsigned Lebesgue integral as the lower unsigned lebesgue integral, with no mention of the upper lebesgue integral. So according to the definition choosed the Lebesgue unsigned integral exists even if the lower and upper integral are not equal.

  • $\begingroup$ Just to make sure - if $f$ is Riemann integrable, then its Lebesgue integral is equal to Riemann integral, and also its upper Lebesgue integral? $\endgroup$ – user4205580 Nov 11 '15 at 21:13
  • $\begingroup$ Yes, because riemann integrable means that your function has a compact support and is bounded. $\endgroup$ – Renart Nov 12 '15 at 15:53

The upper and lower integrals can disagree on a set of Lebesgue measure 0, i.e "$\lambda$ - a. e."

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    $\begingroup$ Some more detail would improve this answer! $\endgroup$ – 6005 Oct 29 '15 at 18:25

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