Convergence in probability implies weak convergence to a Dirac Delta I am trying to show that, for $c \in \mathbb{R}^d$ constant, if $X_n \rightarrow^{\mathbb{P}} c$, then $\mathbb{P}^{X_n} \Rightarrow \delta_c$, where $X_n:\Omega \rightarrow \mathbb{R}^d$ is a random variable.
My attempt was seeing that, since $X_n \rightarrow^\mathbb{P} c$, then, for continous (and bounded) function $f$ in $\mathbb{R}^d$ and a subsequence $\{X_{n_k}\}$, we have that $f(X_{n_k}) \rightarrow f(c)$ a.e. Since $f$ is bounded, the DCT gives that $\int f(X_{n_k})d\mathbb{P}^{X_n} \rightarrow \int f(c) d\mathbb{P}^{X_n} = f(c)$. Since we could also write $f(c) = \int f d\delta_c$, the proof is concluded.
I really don't know why, but it does not seem to be right. Could someone give me a hand?
 A: The other answer is true, but I think it's good to have a more fundamental answer that uses only the concepts in the question to get to the same answer. Also the answer I give uses literally everything we are given to arrive at the conclusion, which I find satisfying. Also, as a notational choice, I'm going to use $|\cdot|$ to mean both absolute value and the Euclidean norm.
Proof Fix a continuous and bounded $f$. First note that $\mathbb{E}_{Y \sim \delta_c} [f(Y)] = f(c)$. We will show that for any $\epsilon > 0$ there is an $N$ such that for any $n \geq N$ we have
\begin{equation}
   |\mathbb{E}f(X_n) - \mathbb{E}_{Y \sim \delta_c}(Y)| = |\mathbb{E}f(X_n) - f(c)| < \epsilon
\end{equation}
which implies
\begin{equation}
    \lim \mathbb{E}f(X_n) = \mathbb{E}_{Y \sim \delta_c}f(Y)
\end{equation}
Since this will hold for all $f$ it will imply $X_n$ converge weakly to $Y$.
By the continuity of $f$ we know that there is a $\delta > 0$ such that for all $x$ satisfying $|x - c| < \delta$ we have $|f(x) - f(c)| < \epsilon/2$. Applying Jensen's inequality, we can say that
\begin{align}
    |\mathbb{E}f(X_n) - f(c)| &\leq \mathbb{E}\mathbf{1}_{|X_n - c| < \delta}|f(X_n) - f(c)| \\
    &\hspace{0.5cm} + \mathbb{E}\mathbf{1}_{|X_n - c| \geq \delta}|f(X_n) - f(c)|.
\end{align}
In the first expectation we can use the continuity of $f$
\begin{equation}
\mathbb{E}\mathbf{1}_{|X_n - c| < \delta}|f(X_n) - f(c)| \leq \mathbb{E}\mathbf{1}_{|X_n - c| < \delta}(\epsilon/2) \leq \epsilon / 2
\end{equation}
and in the second we can use the boundedness of $f$
\begin{equation}
\mathbb{E}\mathbf{1}_{|X_n - c| \geq \delta}|f(X_n) - f(c)| \leq \mathbb{E}\mathbf{1}_{|X_n - c| \geq \delta}(2M) = 2M\mathbb{P}\{|X_n - c| \geq \delta\}
\end{equation}
To control the last probability we can use that $X_n$ converges in probability to $c$ and therefore there exists an $N$ such that for $n \geq N$ we have
\begin{equation}
\mathbb{P}\{|X_n - c| \geq \delta \} < \epsilon/4M.
\end{equation}
Using this we have for all $n \geq N$ that
\begin{equation}
    2M\mathbb{P}\{|X_n - c| \geq \delta\} < 2M(\epsilon/4M) = \epsilon/2.
\end{equation}
Summing up we have for $n$ larger than $N$
\begin{equation}
|\mathbb{E}f(X_n) - f(c)| < \epsilon/2 + \epsilon/2 = \epsilon.
\end{equation}
As discussed above, since $\epsilon$ and $f$ were arbitrary this shows that $\mathbb{P}^{X_n}$ converge weakly to $\delta_c$.
Intuition Essentially the main idea is to break things up into two cases. When $X_n$ is near $c$ use the continuity of $f$ to say that $|f(X_n) - f(c)|$ is very small. We can use the convergence in probability to control the probability that $X_n$ is far from $c$, and the boundedness of $f$ gives us at most a fixed constant ($2M$) which the small probability needs to fight against. From there it's just a matter of choosing constants right to make the calculation work out.
Further Point What if we didn't have the boundedness assumption on $f$? Without the boundedness one could cook up the case
$\mathbb{P}(X_n = 0) = 1-1/n$ and $\mathbb{P}(X_n = n^2) = 1/n$ and consider $f(x) = x$. Then $X_n$ would converge in probability to $\delta_0$ but
$$\lim_{n\rightarrow \infty} \mathbb{E}f(X_n) = \lim_{n\rightarrow \infty} (1-1/n)0 + (1/n)n^2 = \lim_{n \rightarrow \infty} n = \infty$$
which disagrees with $\mathbb{E}_{Y \sim \delta_0} f(Y) = 0.$
A: You just proved that there exists a subsequence $(X_{n_k})$ such that $\mathbb P^{X_{n_k}}\implies\delta_c$.
What you want to prove is the very well known fact that convergence in probability implies convergence in distribution. This is also true if the limit is not a deterministic constant $c$ but a random variable $X$.
There are a lot of ways to prove this. If we want to stay in the spirit of your proof, take any subsequence $(X_{n_k})$. Then there exists a sub-subsequence $(X_{n_{k_{p}}})$ such that $X_{n_{k_{p}}}$ converges almost surely to $c$. Therefore $\mathbb P^{X_{n_{k_{p}}}}\implies\delta_c$. This proves that $\mathbb P^{X_n}\implies\delta_c$.
Note: I used the fact that if a convergence is implied by a topology, then a sequence $(x_n)$ converges to x iff from any subsequence $(x_{n_k})$ we can find a sub-subsequence $(x_{n_{k_p}})$ which converges to $x$.
