I am looking at the proof of the Ito's formula given in Rene Schilling's Brownian Motion.

Theorem. Let $B_t$ be a one dimensional Brownian motion and let $f:\mathbb{R}\to \mathbb{R}$ be a $C^2$-function. Then we have for all $t \ge 0$ almost surely $$f(B_t)-f(B_0) = \int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^t f''(B_s)ds.$$

The proof proceeds as first assuming that the support of $f$ is compact, and then showing that for any partition $t_0=0<t_1<\cdots <t_N=t$ we have the decomposition $$f(B_t)-f(B_0) = \sum_{l=1}^N f'(B_{t_{l-1}})(B_{t_l}-B_{t_{l-1}})+\frac{1}{2} \sum_{l=1}^N f''(\xi_l)(B_{t_l}-B_{t_{l-1}})^2 =:J_1 + J_2$$ where $\xi_l=\xi_l(\omega)$ is an intermediate point between $B_{t_{l-1}}$ and $B_{t_l}$.

The proof is complete by showing that $J_1 \to \int_0^t f'(B_s)dB_s$ in probability as $|\Pi|\to 0$ and $J_2\to \frac{1}{2}\int_0^t f''(B_s)ds$ in probability as $|\Pi|\to 0$. Finally, we use cutoff functions in the general case and using the above identity for $t\wedge \tau(l)$, where $\tau(l):=\inf \{s>0: |B_s|\ge l\}$, use $\lim_{l\to \infty} \tau(l)=\infty $ almost surely to complete.

Question: In the theorem, how do we get the identity almost surely? The proof actually shows that the LHS in the Ito's formula converges in probability to the RHS. How does this imply that the identity holds a.s.?

I.e. the proof shows that $P-\lim_{|\Pi|\to 0} (f(B_t)-f(B_0))=\int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^t f''(B_s)ds$, but how does this mean that $f(B_t)-f(B_0)= \int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^t f''(B_s)ds$?

  • $\begingroup$ thanks for asking this question! I was really unsure how to look it up online and fortunately enough you were the first result :) $\endgroup$ Aug 17, 2021 at 15:31

2 Answers 2


I think there is a confusion here on which quantity converges to which quantity and in what sense. The second assertion regarding almost sure convergence is about : $$\frac{1}{2}\int_0^{t\wedge \tau(l)} f''(B_s)ds\to \frac{1}{2}\int_0^t f''(B_s)ds $$


$$\int_0^{t\wedge \tau(l)} f'(B_s)dB_s\to \int_0^t f'(B_s)dB_s$$

not on the limit of : $$P-\lim_{|\Pi|\to 0} (f(B_t)-f(B_0))=\int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^t f''(B_s)ds$$ being replaced by almost sure convergence because as you may know (naive or deterministic) stochastic integration cannot be defined in an almost sure way (have a look at the first chapter section 8 of Protter's book)(*).

So now almost sure convergence is trivial for $\frac{1}{2}\int_0^{t\wedge \tau(l)} f''(B_s)ds$ as the almost sure convergence of $t\wedge \tau(l) \to t$ is only property from usual integration seen at a fixed $\omega$.($ \int_0^{t_n(\omega)} f(s)ds \to \int_0^t f(s)ds$ as $t_n(\omega) \to t$ almost surely)

And for the second you have to use quadratic variation to show the same point because two processes with the same quadratic variation (path by path) are equal up to a constant which is not too hard to show here to be null.

(*) Speaking in full generality it depends on what properties you want for your stochastic integral, there are ways to almost sure stochastic integration but you have to lose something, anyway in the Schilling's book context those considerations are irrelevant IMO.

  • $\begingroup$ Ok, so in the bounded functional case, I think what the proof shows is that $f(B_t)-f(B_0)$ converges in probability (as mesh of $\Pi$ goes to $0$) to the sum of stochastic integrals $\int_0^t f'(B_s)dB_s + 1/2 \int_0^t f''(B_s)ds$, but then $f(B_t)-f(B_0)$ does not depend on the partition $\Pi$, so it also converges in probability to $f(B_t)-f(B_0)$, i.e. itself, and hence we have the a.s. equality of two of the limits. And then we use your argument in the general case. Is this interpretation correct? $\endgroup$ Dec 22, 2017 at 13:12
  • $\begingroup$ Well you have $f(B_{t \wedge \tau})-f(B_0)=\int_0^{ t \wedge \tau} f'(B_s)dB_s + 1/2 \int_0^{t \wedge \tau} f''(B_s)ds \to \int_0^t f'(B_s)dB_s + 1/2 \int_0^t f''(B_s)ds $ almost surely and $f(B_{t\wedge \tau})-f(B_0)\to f(B_t)-f(B_0)$ almost surely so that both limits match if this is what you mean in your comment that's right. Regards $\endgroup$
    – TheBridge
    Jan 4, 2018 at 12:12

As TheBridge has said, Itô integral as a process is ambiguous on a null set. Nevertheless you can always find a martingale equals to the Itô integral almost everywhere. Note that if you have a sequence coverages in probability you can find a subsequence converges almost surely. So you can prove the almost sure limits of subsequence $J_1,{n_k}$ and $J_2,{n_k}$ equal to the RHS.


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