Conditions under which pointwise limits of functions are Riemann integrable. A linchpin of Lebesgue theory is that limits of measurable functions are measurable.

Is there any similar condition for Riemann integrability? Can we create a class of functions for Riemann integrability that is closed under limits?

One class could be "the set of continuous functions under uniform limits". But can we modify this in some way to allow pointwise limits?

From what I remember there is a proof possible using classical (Riemann integrable arguments) that you can do which say if your limit function is Riemann integrable, then the integral and the limit commute (this is obviously true as it's a consequence of Lebesgue theory).

I think one set would be equicontinuous functions using some fixed delta, any others which are more relaxed?
It is definitely not true that the pointwise limit of Riemann integrable functions is Riemann integrable. 

Let $\mathbb{Q}= \{q_n: n\in\mathbb{N}\}$ be the rationals. Then, define;
\begin{equation}
f_N(x)= \sum_{n=1}^n\chi_{q_n}(x),
\end{equation}
where,
\begin{equation}
\chi_{q_n}(x)= \begin{cases}
1, x= q_n\\ 
0, x\neq q_n.
\end{cases}
\end{equation}

It is clear that each $f_N$ is Riemann integrable (use definition) with $\int f_N= 0$. However, the pointwise limit, $f_N\to \chi_{\mathbb{Q}}$ is not Riemann integrable.
$y=sinx$ between $0$ and $\pi$ is Riemann integrable. So are the partial sums of Fourier series between the same limits. A Fourier series can be found for a step function. The step function is Riemann integrable, but the limit of a sequence of step functions might not be. So even "nice" functions can generate functions that are not Riemann integrable. This lack of "closure" of Riemann integrals showed the need for Lebesgue theory.