Let $X_1,\ldots,X_n\mid \theta\stackrel{i.i.d}{\sim}N(\theta,\sigma^2),\ \theta\in\mathbf{R},\ $ and $\ \sigma^2 > 0$ is known.
Let $\pi_n(\theta\mid x_n)$ be the posterior distribution when the prior distribution is $\pi_n = N(0, n), n \in\mathbf N$
and $\pi(\theta\mid x_n)$ be the posterior distribution when the (improper) prior distribution is uniform.
Show that $$ ||\pi_n(\theta\mid x_n) - \pi_n(\theta\mid x_n)||\stackrel{p}{\longrightarrow} 0 $$
Here, $||f-g||_1=\int|f(x)-g(x)|\,dx$.
My hopeless attempt
$$\pi_{n}\left(\theta \mid x_{n}\right) \propto e^{-\frac{1}{2n} \theta^{2}} \prod_{i=1}^{n} e^{-\frac{1}{2 \theta^{2}}\left(x_{i}-\theta\right)^{2}}\propto e^{-\frac{1}{2}\left(\frac{1}{n}+\frac{n}{\sigma^{2}}\right)\left(\theta-\frac{\frac{n}{\sigma^{2}} \bar{x}}{\frac{1}{n}+\frac{n}{\theta^{2}}}\right)^{2}}$$
Therefore, $$\theta_{\left(\pi_{n}\right)} \mid x_{n} \sim N\left(\frac{\frac{n}{\sigma^{2}} \bar{x}}{\frac{1}{n}+\frac{n}{\sigma^{2}}}, \frac{1}{\frac{1}{n}+\frac{n}{\sigma^{2}}}\right)$$
$$\pi\left(\theta \mid x_{n}\right) \propto 1\cdot\prod_{i=1}^{n} e^{-\frac{1}{2 \sigma^{2}}\left(x_{i}-\theta\right)^{2}}\propto e^{-\frac{1}{2\left(\frac{\sigma^2}{n}\right)}(\theta-\bar{x})^{2}}$$
Therefore, $$\theta_{(\pi)} \mid x_{n} \sim N\left(\bar{x}, \frac{\sigma^{2}}{n}\right)$$
Let $\ \mu_{n}=\frac{\frac{n}{\sigma} \bar{x}}{\frac{1}{n}+\frac{n}{\sigma^{2}}}, \quad \sigma_{n}^{2}=\frac{1}{\frac{1}{n}+\frac{n}{\sigma^{2}}}$
Then
$\begin{aligned} & \int\left|\pi_{n}\left(\theta \mid x_{n}\right)-\pi\left(\theta \mid x_{n}\right)\right| d \theta \\=& \int \left|\left(2 \pi \sigma_{n}^{2}\right)^{-\frac{1}{2}} \exp \left(-\frac{1}{2}\left(\frac{\theta-\mu_{n}}{\sigma_{n}}\right)^{2}\right)-\left(2 \pi \frac{\sigma^{2}}{n}\right)^{-\frac{1}{2}} \exp \left(-\frac{1}{2}\left(\frac{\theta-\bar{x}}{\frac{\sigma}{\sqrt{n}}}\right)^{2}\right) \right| d \theta \end{aligned}$
...
I have no idea how I could derive convergence in probability from here.