Let $f$ be a continuous function, and $B_t$ be a Brownian motion starting at $0$ at time $t=0$ on the filtered probability space $(\Omega, \mathcal{F}, P)$.
I am trying to establish (or find a contradiction to the claims) that $f(B_t)$ is adapted to the filtration $\mathcal{F}_t$ generated by $B_t$ and that $E[\int_{0}^{T}f(B_s)^2ds]<\infty$ P-a.s.
To show that $f(B_t)$ is adapted, i need to show that $f(B_t)$ is $\mathcal{F}_t$-measurable for all $t$. This is to show that for each borel set $B \in \mathbb{B}(\mathbb{R})$ we have \begin{align} \{\omega \in \Omega \ | \ f(B_t) \in B\} \in \mathcal{F_t}. \end{align}
The argument I have in mind for this part, is to use the fact that $(f\circ B_t)^{-1} = B_t^{-1} \circ f^{-1}$. Since we have that for each Borel set $B$, $f^{-1}(B)$ is an open set and hence itself a Borel set, the measurability follows by the measurability of $B_t$. This is to say,
\begin{align} \forall B \in \mathbb{B}(\mathbb{R}), \ \{\omega \in \Omega \ | \ f(B_t) \in B\} \in \mathcal{F_t}. \end{align}
However, I am not finding a way with the square integrability part. I had hoped to use the fact that the brownian motion has continuous paths P-a.s to bound $|\sup_{s \in [0, t]}(B_s)(\omega') -\inf_{s \in [0, t]}(B_s)(\omega')|$ for fixed omega. Then, if this holds for almost all $\omega'$, perhaps I can show that $E[\int_{0}^{T}f(B_s)^2ds]<\infty$?
Perhaps the arguemnt is not sound or helpful?
Unfortunately it's false.
Consider $f: \mathbb{R}\to \mathbb{R}$ defined as $f(x) = e^{\frac{ax^2}{2}}$ for a certain constant $a>0$.
Then, $\mathbb{E}[f(B_s)^2] = \mathbb{E}[e^{B_s^2}] = \int_\mathbb{R} \frac{1}{\sqrt{2\pi}}e^{asx^2-\frac{x^2}{2}} dx$, this is $+\infty$ if $s\geq \frac{1}{2a}$.
So, $\mathbb{E}[\int_0^Tf(B_s)^2ds] = \int_0^T \mathbb{E}[f(B_s)^2] ds = +\infty$ whenever $T>\frac{1}{2a}$.
However, as you mentioned, you can fairly easily (using the continuity of the Brownian Motion) show that $\mathbb{P}(\int_0^Tf(B_s)^2ds = +\infty) = 0$.