What are the advantages of the Riemann vs. Lebesgue Integral? I am quite new to analysis and I have a basic question. Whenever the Riemann integral is defined, the Lebesgue integral is defined and the two are the same. Perhaps the Riemann integral is easier to construct. Besides this, are there any theoretical reasons to consider the Riemann Integral instead of the Lebesgue?
For instance, what is an important property that the Riemann integrable functions have?
I'm also wondering why it is the Riemann integral that is predominantly taught and not the Lebesgue.
 A: You asked for an important property of Riemann integrable functions.

A bounded function defined on $[a,b]$ is Riemann integrable iff the set of its points of discontinuity has measure zero.

Note functions like $\mathbb{1}_\mathbb{Q}$ do not meet this criterion (for nontrivial intervals), but are still Lebesgue integrable. The criterion for a bounded function on $[a,b]$ to be Lebesgue integrable is that the function be measurable, which requires defining a measure, which takes you on a medium-sized detour.* 
Lebesgue integration applies to strictly more functions than Riemann does**. And all the same limiting processes are possible: two-sided improper, or requiring that the limit from either side exist individually, including when the point is $\infty$.
Why use the Riemann integral at all? It's a good example of a mathematical construction through a limiting process, which allows teachers to open the black box of the integral without undue pain. It arguably is the more intuitive definition.

*-"[Lebesgue] measure zero" is easier to explain than an entire measure.
**-There is an exception which is an artifact of the definitions. For an integral like 
$$\int_0^\infty \frac{\sin x}{x}$$
the convention for Riemann integration would be to define the limit as 
$$\lim_{t \to \infty} \int_0^t \frac{\sin x}{x}.$$
If we accept this definition of the limit, the limit exists for Riemann or for Lebesgue. In the Lebesgue integral, it is more usual to place a more stringent requirement on the function, namely that $\int_0^\infty \max(f, 0)$ and $\int_0^\infty -\min(f, 0)$ exist separately. In this sense both Riemann and Lebesgue are divergent. But because the Lebesgue integral mostly uses the second definition, one can say somewhat misleadingly that "the improper Riemann integral exists but not the improper Lebesgue integral". 
