So, I'm battling with two problems:
Show that:
$$ \int_0^1 \frac{B_s}{s} ds $$ is finite
$$ B_t - \int_0^t \frac{B_s}{s} ds $$ is Brownian Motion
My attempt was to start with: $$ d(B_t \ln t) = \ln (t) dB_t + \frac{B_t}{t} dt $$.
Then we can write: $$ B_1 \ln(1) - B_0 \ln(0) = \int_0^1 \ln(s) dB_s + \int_0^1 \frac{B_s}{s} ds$$
Left handside is equal to zero, assuming $0 \ln 0 = 0$. First term on the right is integrable, since $\int \ln^2(s) ds < \infty$, so it's martingale. Then the one that interests us is integrable as well.
This approach seems rather fishy, at best, and straight out wrong, at worst, but I don't have any better idea right now.
Second one: $$ \mathbb{E} B_t \vert \mathcal{F}_s = B_s$$
$ \int_0^l B_s/s ds $ is $\mathcal{F}_l$-measurable. Then $\mathbb{E} \int_l^t B_s / s ds = \int_l^t \mathbb{E} B_s /s ds = 0$, so whole expression is martingale with quadratic variation equal to $t$, because second part is finite variation process, hence it's BM.
However, I have a feeling that it should be doable without invoking Levy's characterization. How?
So, my questions, as usual - is any of these close to being correct? If not, what is correct approach and where are the mistakes?