My definition of the stochastic integral is that it it is the image of the Ito isometry. Now we also prove Ito's formula and then apply it pathwise and get a pathwise definition in some cases. But in Ito's formula the following definition of the stochastic integral is used:
$$\int_0^t f(X_s) dX_s = \lim_{n\to\infty} \sum_{s \in \pi_n} f(X_s) (X_{s'\wedge t} - X_{s\wedge t})$$
Where $\pi_n$ is a partition of $\mathbb{R}$ with $\text{mesh}(\pi_n) \to 0$ and $s'$ is the successor of $s$ in $\pi_n$.
The right-hand side is the Riemann-sum approximation of the integral. But we only have proven that this converges to the integral in some special cases ($f(X_s)$ continuous and bounded for example). The only condition on $f$ is that it lies in $C^1(\mathbb{R})$ here.
I understood that the right-hand-side converges - but why should it converge to the stochastic integral defined by the Ito isometry?
Edit: To elaborate on my definition of the stoch. integral: We have defined the stochastic integral for the simple processes $E$ which are of the kind $H := \sum_{i=0}^n 1_{(a_i, a_{i+1}]} A_i$ with $A_i$ being $F_{a_i}$-measureable and bounded. For this kind of processes the stochastic integral is defined as
$$J(H) := \int_0^t H_t dM_t = \sum_{i=0}^n A_i (M_{a_{i+1}} - M_{a_i})$$
Now the ito isometry tells us that $||H||_{L^2(\Omega \times [0, u), \mathbb{P} \times \lambda)} = ||J(H)||_{M^2([0,u))}$ where the hilbert space $M^2([0,u))$ consist of equivalence classes of square-integrable martingales, $M \sim M'$ iff $M$ and $M'$ are modifications of each other, i.e. $\forall t: \mathbb{P}(M_t = M'_t) = 1 $. Now $J$ implies a unique isometry $\overline{J}$ on the closure of $E$ in $L^2(\Omega \times [0, u)$.
$\overline{J}(H)$ for $H \in \overline{E}$ is the stochastic integral.