I am pondering over following question:
Let $B_t$ be a standard Brownian Motion. Is the following Itô-Integral well-defined?
$$ X_{\infty} = \int\limits_1^{\infty} B_{\frac{1}{t}} \,dB_t $$
My Attempt:
When considering the same integral $X_T$ with a finite bound $T>1$, then the integral is well-defined as $f(t)=B_{\frac{1}{t}}$ is progressivly measureable (see addition at the end) and $E[\int_1^T (B_{\frac{1}{t}})^2 \,dt] = \ln(T) < \infty$ via Fubini-Tonelli.
If you then take $T\rightarrow \infty$, then $E[X_{\infty}]=0$ and $E[\int_1^T (B_{\frac{1}{t}})^2 \,dt] = \infty$, thus $X_{\infty}=0$ a.s. and $X_T$ is well-defined.
Is my reasoning correct or did I make a mistake somewhere??
Additional question: Is it enough for an Itô-Integral to be well defined with infinite upper bound, if it is well-defined for alle finite upper bounds??
Thank you in advance.
Addition to progressivly measureability of $f(t)=B_{\frac{1}{t}}$:
Consider the natural filtration $\mathcal{F}_t = \sigma(B_s: s\leq t)$, then $(B_{\frac{1}{t}})_{1\leq t < \infty} = (B_s)_{0 < s \leq 1}$ is $\mathcal{F}_1$ (and thus $\mathcal{F}_T$ for all $T>1$) measureable.
As @KurtG. commented, in order for the integral to be well defined it is required that $\int_1^\infty B_{1/t}^2 dt<\infty$ almost surely. I'll show that, vice versa, it is infinite almost surely.
First we apply the well-known fact that $(B_{1/t},t\ge 0)$ has the same distribution as $(B_{t}/t, t\ge 0)$, so it is equivalent to consider $\int_1^\infty \frac{B_t^2}{t^2}dt$ (actually, this step may be skipped, but in my opinion this expression is nicer).
Consider the sequence $$ X_n = \int_{2^{n-1}}^{2^{n}} \frac{B_t^2 }{t^2}dt. $$ It is easily seen to be strictly stationary due to self-similarity of Brownian motion. Therefore, by the ergodic theorem, $\frac 1n \sum_{k=1}^n X_k$ converges almost surely to some positive random variable$^*$, in particular, $$\int_1^{2^n} \frac{B_t^2}{t^2}dt = \sum_{k=1}^n X_k \to +\infty, \quad n\to\infty,$$ almost surely.
$^*$Actually, to a positive constant, since $X$ is also ergodic.