If $\int_0^T f(t) \, dW_t=0$ then $f=0$

Question

Let $f$ be a real-valued stochastic process adapted to the natural filtration $\{\mathcal{F}_t\}_{t\geq0}$ of some Brownian motion $\{W_t\}_{t\geq0}$ such that

$\mathbb{E}\bigg[\int_{0}^{T}f^2(t,w)dt\bigg]<\infty $ and $\int_{0}^{T}f(t,w)dW_t=0.$

Then how one can prove that $f:=0$ for all $t\in[0,T].$

By Ito isometry, $\mathbb{E}\bigg[\int_{0}^{T}f^2(t,w)dt\bigg]=\mathbb{E}\bigg[\int_{0}^{T}f(t,w)dW_t\bigg]^2=0$. This implies $\mathbb{E}\bigg[\int_{0}^{T}f^2(t,w)dt\bigg]=0$. Then I got stuck to proceed. Anyone help?

Answer 1

If $\mu$ is a non-negative measure on a measure space $(X,\mathcal{A})$ and if $g\ge0$, then $ \int_X g(x) \, d\mu(x) = 0$ implies $\mu(g \neq 0)=0$.

Here, we consider the product measure $\mu := \lambda_T\otimes \mathbb{P}$ (where $\lambda_T $ denotes the Lebesgue measure restricted to $[0,T]$). Since, by Tonelli's theorem,

$$0=\mathbb{E} \left( \int_0^T f^2(t,\cdot) \, dt \right)= \int_{[0,T] \times \Omega } f^2(t,\omega) \, d\mu(t,\omega)$$

we find $$\mu(\{(t,\omega) \in [0,T] \times \Omega; f(t,\omega) \neq 0\})=0,$$ i.e. $f=0$ for almost all $t,\omega$.