Here is an exercise from Rudin's "Real and Complex Analysis" which I think shows the point:
If $\{ f_n \}$ is a sequence of continuous functions on $[0,1]$ such that
$0 \leq f_n \leq 1$ and such that $f_n(x) \to 0$ as $n \to \infty$ for every
$x \in [0,1]$, then
$$ \lim_{n \to \infty} \int_0^1 f_n(x) dx = 0$$
Try to prove this without using any measure theory or any theorems about
Lebesgue integration. (This is to impress you with the power of the Lebesgue
integral. A nice proof was given by W.F. Eberlein in Communications on Pure and
Applied mathematics, vol. X, pp. 357-360, 1957.)
The solution involving the Dominated Convergence Theorem is as follows:
Notice the functions $f_n$ are dominated by the constant $1$ function, which is $L^1$ since $[0,1]$ has finite measure. Then we see
$$ \lim_{n \to \infty} \int_0^1 f_n(x)~d\mu \overset{(1)}{=} \int_0^1 \lim_{n \to \infty} f_n(x)~d\mu \overset{(2)}{=} \int_0^1 0~d\mu = 0 $$
where (1) follows from the DCT and (2) follows from the assumption $f_n(x) \to 0$. If you want to be extremely precise, you can then note that the $f_n$ and $0$ are all riemann integrable too. Since lebesgue integration and riemann integration coincide for riemann integrable functions, this means we have also shown the claim with $d\mu$ replaced by $dx$ above... though I feel this is a bit of a pedantic note.
The important part of the exercise, though, is the following: Try to prove this theorem without the machinery of measure theory. I certainly don't know how I would do it, and that's the point.
In general, the lebesgue integral works better with respect to limits because of the Dominated Convergence Theorem and its sister the Monotone Convergence Theorem. Since we are often interested in taking limits, this is obviously a benefit. In fact, this benefit is even larger than it initially seems:
The space of Riemann integrable functions on $[0,1]$, with its natural inner product
$$\langle f, g \rangle = \int_0^1 f(x) \cdot g(x) dx$$
is not complete, because there are Cauchy sequences of riemann integrable functions whose limit is not riemann integrable (exercise :P). The lebesgue integral solves this problem (using the DCT), allowing for the use of Banach and Hilbert Space theory.
I hope this helps ^_^
Best Answer
Since Lebesgue's integral extend Riemann's, every convergence theorem for Lebesgue integral also holds for Riemann integral as long as the sequence of functions involved, and its limit, are Riemann integrable functions.
However, the pointwise limit of Riemann integrable functions might not be Riemann integrable, so stating such results in the context of Riemann integral would look a bit awkward.
Moreover, as I believe underlies @Kavi's comment, Riemann integration is better suited to introducing students to the theory of integration at an early stage, when they might not yet be prepared to work with so many different notions of convergence, so I think it is pedagogically correct to stick with the simplest convergence Theorem, namely the one about uniform convergence.