Let $X_1,X_2,...,X_n | \theta \sim N(\theta,\sigma^2), \theta \in \mathbb{R}, \sigma^2>0,$ and if prior distribution is $\pi_n = N(0,n) , n\in \mathbb{N} $, then posterior distribution is $\pi_n(\theta|X_1,...X_n)$ and if prior distribution is uniform distribution $U(a,b)$, then posterior distribution is $\pi(\theta|X_1,...,X_n)$. Show
$$\|\pi_n(\theta|X_1,...,X_n)-\pi(\theta|X_1,...,X_n)\|_1 \rightarrow 0 \;\; in \;\; probability$$
where $\|f-g\|_1 := \int |f(x)-g(x)|dx,$ $f$ and $g$ are probability density functions which defined on $\mathbb{R}$.
How do we prove that convergence?
I only show that
$$ \begin{array}{} \pi_n(\theta | X_1,...,X_n) &\propto \prod_{i=1}^n P(X_i|\theta) \cdot \pi_n(\theta)\\ &= (\frac{1}{\sqrt{2\pi}\sigma})^n \cdot exp(-\frac{\sum_{i=1}^n(X_i-\theta)^2}{2\sigma^2})\cdot\frac{1}{\sqrt{2\pi n}} \cdot exp(-\frac{\theta^2}{2n})\\ &\propto exp(-\frac{(\sigma^2+n^2)(\theta-n\sum X_i/(\sigma^2+n^2))^2}{2n\sigma^2})\\ &\propto N(\frac{n\sum X_i}{\sigma^2+n^2}, \frac{n\sigma^2}{\sigma^2 +n^2}) \\ \\ \\ \pi(\theta |X_1,...X_n) &\propto exp(-\frac{\sum (X_i-\theta)^2}{2\sigma^2})\cdot I(a<\theta<b) \end{array} $$ .