This is a proof verification question. The claim that I want to prove is as follows.
Let $(B,W)$ be two dimensional Brownian motion. Define the stopping time $\sigma_{n}$ as $$\sigma_n = \inf\left\{t > \frac{1}{n}: B_t = 0\right\}$$ Then, $W_{\sigma_n}$ has a probability density.
My proof: $\sigma_{n}$ and $W$ are independent. Therefore, I can write $$E\left[e^{i\theta W_{\sigma_n}}\right] = E\left[e^{-\frac{1}{2}\theta^2\sigma_{n}}\right]$$ for any $\theta \in \mathbb{R}$. Next, I want to employ the Fourier inversion theorem. \begin{align} \int_{\mathbb{R}}E\left[e^{-\frac{1}{2}\theta^2\sigma_{n}}\right]\,d\theta &= E\left[\int_{\mathbb{R}}e^{-\frac{1}{2}\theta^2\sigma_{n}}\,d\theta \right]\\ &\leq E\left[\int_{\mathbb{R}}e^{-\frac{1}{2n}\theta^2}\,d\theta \right] \end{align} In going from the first row to the second I made use of the fact that $\sigma_n \geq \frac{1}{n}$. Since the last expression is finite, the characteristic function of $W_{\sigma_n}$ is integrable. Hence, it possesses a density.
Is this proof correct? If not, how can I correct it?