\begin{proof} Recall the Lebesgue Integral Triangle Inequality: Lebesgue Integral Triangle Inequality this helps us find: $$\begin{align*} \sup_{A\in\mathcal{F}}|\nu(A)-\nu_{n}(A)|&= \sup_{A\in\mathcal{F}}\left| \int\limits _{A}\delta \, d\mu-\int\limits _{A}\delta_{n} \, d\mu \right| \\ &= \sup_{A\in\mathcal{F}}\left| \int\limits _{A}\delta-\delta_{n} \, d\mu \right| \\ &\le \sup_{A\in\mathcal{F}}\int\limits _{A}|\delta-\delta_{n}| \, d\mu \end{align*}$$then, let $g_{n}=\delta-\delta_{n}$.

The positive part $g_{n}^{+}$ of $g_{n}$ converges to 0 a.e. (this is trivial).

Since:$$0\le g_{n}^{+}\le \delta \iff 0\le(\delta-\delta_{n})^{+}\le\delta$$and $\delta \in\mathscr{L}^{1}(\Omega,\mathcal{F},\mu)$ and so Dominated Convergence Theorem applies: $$\int\limits g_{n}^{+} \, d\mu\to0.$$But, $\int\limits g_{n} \, d\mu=0$ by $(1)$ hence $$\begin{align*} \int\limits _{\Omega}|g_{n}| \, d\mu &= \int\limits _{\{ g_{n}\ge0 \}}g_{n} \, d\mu-\int\limits _{\{ g_{n}<0 \}}g_{n} \, d\mu\\ &= 2\int\limits _{\{ g_{n}\ge 0\}}g_{n} \, d\mu\\ &= 2\int\limits _{\Omega}g_{n}^{+} \, d\mu\to {0} \end{align*}$$ where the third line arises since the integral is equal to zero.

\end{proof} This theorem shows up in a different form in Borkar’s Probability Theory: An Advanced Course: >[!thm|2.33] Scheffé’s Theorem >Let $(\mu_{n})_{n\in\mathbb{N}}$ be a sequence of probability measures on a measurable space $(\Omega,\mathcal{F})$ such that $$\mu_{n}\ll\lambda,\quad \forall n\in\mathbb{N},$$for some nonnegative, σ-finite measure $\lambda$ on $(\Omega,\mathcal{F})$. If $$\frac{d\mu_{n}}{d\lambda}\to \frac{d\mu_{\infty}}{d\lambda},\,\lambda \text{-a.s.}$$then $\mu_{n}\to \mu_{\infty}$ in total variation (which is equivalent to Convergence in Distribution here).

\begin{proof} The claim amounts to showing that $$\frac{d\mu_{n}}{d\lambda}\to \frac{d\mu_{\infty}}{d\lambda},\,\text{ in }L^{1}(\Omega,\mathcal{F},\lambda)$$this follows from the following theorem: >[!thm|1.3.3] >Let $(X_{n})_{n\ge 1}$ be a sequence of integrable rvs on $(\Omega,\mathcal{F},\mathbb{P})$ with $X_{n}\ge 0,\forall n\in\mathbb{N}$ a.s. and $X_{n}\xrightarrow{\text{a.s.}}X$. Then $$X_{n}\xrightarrow{L^{1}}X \iff \mathbb{E}[X_{n}]\to \mathbb{E}[X]$$ >

For $S$ countable with the discrete topology, this shows that convergence in $\mathcal{P}(S)$ and in total variation are equivalent. \end{proof} >[!remark] >This gives us $$f_{n}\xrightarrow{a.s}f\implies X_{n}\xrightarrow{d}X$$only if $X_{n}\ll X$.