I know that for $f:[0,1]\to \mathbb{R}$, the Riemann Integral converges in the sense that $$\sum_{k=1}^Kf(t_k)(t_{k} - t_{k-1})\longrightarrow \int_0^1f(t)dt$$ as the grid becomes smaller and smaller.
For some diffusion process $X_t$ and some function $f$, does $$\sum_{k=1}^Kf(t_k)(X_{t_k} - X_{t_{k=1}}) \longrightarrow\int_0^1f(t)dX_t$$ and in which sense?
The first important thing to notice is that we cannot expect convergence for fixed $\omega$, i.e. the sum
$$\sum_{k=1}^K f(t_k) (X_{t_k}(\omega)-X_{t_{k-1}}(\omega))$$
does in general not converge as $K \to \infty$. Recall the following lemma:
Since a Brownian motion $(B_t)_{t \geq 0}$ has infinite variation, this lemma shows that the Riemann sums
$$\sum_k f(t_k) (B_{t_k}(\omega)-B_{t_{k-1}}(\omega))$$
do (in general) not even converge for continuous $f$.
That's why Itô came up with a new idea: Let's consider the $L^2$-limit, i.e. define the stochastic integral as
$$\int_0^T f(t) \, dB_t := L^2-\lim_{K \to \infty} \sum_{k=1}^k f(t_{k-1}) (B_{t_k}-B_{t_{k-1}}).$$
Using martingale techniques, one can show that this limit does indeed exist for a large class of functions $f$ and today that's the usual way to define stochastic integrals with respect to stochastic processes. Note that $L^2$-convergence implies convergence in probability, so we have in particular
$$\sum_{k=1}^k f(t_{k-1}) (B_{t_k}-B_{t_{k-1}}) \stackrel{\mathbb{P}}{\to} \int_0^T f(t) \, dB_t.$$
The so-defined stochastic integral is well-defined for all functions $f:[0,\infty) \times \Omega \to \mathbb{R}$ which are progressively measurable and satisfy the integrability condition
$$\mathbb{E} \left( \int_0^T f(s)^2 \, ds \right) < \infty.$$