I am wondering how to relate the infinitessimal generator of a diffusion process and Ito's formula. Formally, consider the one-dimensional case, and let $dX_t=a(X_t)\,dt+b(X_t)\,dW_t$, where $W_t$ is a standard brownian motion. I know that the infinitessimal generator, for $f$ suitable, is defined as $\mathcal{L}f(X_t)=\lim_{\delta\to 0} \frac{\mathbb{E}^{X_t}[f(X_{t+\delta})-f(X_t)]}{\delta}$.
I know from Oksendal's Book on SDEs that $\mathcal{L}f(X_t)=a(X_t)f'(X_t)+\frac{1}{2}b(X_t)^2f''(X_t)$, and I'm wondering how to reconstruct this formula in a simple way from Ito's formula, by looking at infinitesimal changes (It is usual to express the generator in terms of infinitesimal changes without much justification in finance books).
My attempt is the following. For $\delta$ arbitrarily close to zero, we can think of $\frac{1}{\delta}=\frac{1}{dt}$ and of $f(X_{t+\delta})-f(X_t)=df(X_t)$. Then, by Ito's formula, $$df(X_t)=f'(X_t)\,dX_t+\frac{1}{2}f''(X_t)(dX_t)^2$$
So plugging back $dX_t$ and using Ito rules to compute $(dX_t)^2$ (i.e. $dt\,dW_t=(dt)^2=dW_t\,dt=0$ and $(dW_t)^2=dt)$, we have
$$df(X_t)=f'(X_t)(a(X_t)\,dt+b(X_t)\,dW_t)+\frac{1}{2} f''(X_t)b(X_t)^2 \, dt$$
So taking the expectation, with the observation that $\mathbb{E}^{X_t}(dW_t)=0$, we end up with
$$\mathbb{E}^{X_t}(df(X_t))=f'(X_t)a(X_t) \, dt+\frac{1}{2} f''(X_t)b(X_t)^2 \, dt$$
And thus $\frac{1}{dt}\mathbb{E}^{X_t}(df(X_t))=f'(X_t)a(X_t)+\frac{1}{2} f''(X_t)b(X_t)^2=\mathcal{L}f(X_t)$
My question is the following:
If the reasoning is correct, what do I need to think of $\lim_{\delta\to 0}\frac{\mathbb{E}^{X_t}[f(X_{t+\delta})-f(X_t)]}{\delta}$ as $\frac{\mathbb{E}^{X_t}(df(X_t))}{dt}$? I've seen the latter used mainly in finance books but without much justification. I understand that I need some conditions to make the DCT or MCT work to exchange limit with expectation, but what else would I need to know that $f(X_{t+\delta})-f(X_t)\to df(X_t)$ as $\delta\to 0$? Can someone please help me to fill in the gaps that would make my reasoning completely formal?
