calculus - Lebesgue integral - interpretation

It is often of interest, both in theory and applications, to be able
to pass to the limit under the integral. For instance, a sequence of

functions can frequently be constructed that approximate, in a
suitable sense, the solution to a problem. Then the integral of the
solution function should be the limit of the integrals of the
approximations. However, many functions that can be obtained as limits
are not Riemann integrable, and so such limit theorems do not hold
with the Riemann integral. Therefore, it is of great importance to
have a definition of the integral that allows a wider class of
functions to be integrated (Rudin 1987).

I'm trying to understand the quote above. It says that there are functions $f(x)=\lim_n f_n(x)$ such that $f(x)$ is not Riemann integrable. However one could calculate its Lebesgue integral. Lebesgue integration is powerful because of the Monotone convergence theorem which states that the limit of a sequence of integrals equals integral of the limit (roughly speaking).

Now I'm interested in the following - if Riemann integral of $f(x)=\lim_n f_n(x)$ does not exist, then what meaning should we attach to the number produced by Lebesgue integral? Area under the curve is defined as Riemann integral. What if a function is not Riemann integrable?

If a function is Riemann integrable, then the values of types both integrals are equal - it's perfectly fine. But why should I treat Lebesgue integral as area? Another thing is that with Lebesgue integral, we only give the lower bound of the 'area' under the curve (supremum of simple functions), so basically the upper bound can be completely different. If the upper bound of the area is greater than lower bound, should we really consider Lebesgue integral as 'area'?

One more thing:
If $f$ is a non-negative measurable function on $E$, its Lebesgue integral is defined as
$$\int_E f \, d\mu = \sup\left\{\,\int_E s\, d\mu : 0 \le s \le f,\ s\ \text{simple}\,\right\}$$

then is the following true as well?:

$$\sup\left\{\,\int_E s\, d\mu : 0 \le s \le f,\ s\ \text{simple}\,\right\} =\inf\left\{\,\int_E s\, d\mu : f \ge s \ge \infty,\ s\ \text{simple}\,\right\}$$

Answer

Probably the best way to talk about this is with an example. Let $f_n:[0, 1] \to \mathbb{R}$ where $$ f(x) = \begin{cases} 1 &\mbox{if } x 2^n \in \mathbb{N} \\
0 & \mbox{otherwise. }\end{cases} \pmod{2} $$

Now, you can see that each $f_n$ is integrable, in both the Riemannian and Lebesgue sense. The limit of the sequence is the function that is $1$ for every value with a terminating binary representation and $0$ elsewhere. This is not Riemann integrable as the lower sums and upper sums don't converge to the same value: the lower sums are always $0$ and the upper sums are always $1$, no matter what partition we choose. However, the limit is Lebesgue integrable, with an integral of $0$.

Now, if I were to ask you "what is the area under the curve?" I'd hope that we could agree that its $0$. To paint it I'd have to make a whole lot of lines, but the lines have $0$ width, so... I feel like the area is $0$. Riemann integration just isn't deep enough to see this fact. So, I'd say they're both area, but one is a better, broader idea of area.

Regarding your final question, can the upper bound differ from the lower? Try to relate the two, you should be pleasantly surprised.