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We know that $$\lim_{n\rightarrow\infty}F_n(x)=F(x)$$ whenever $F$ is continuous at $x$ (where $F_n$ and $F$ are the corresponding distribution functions), but it should also be true that $$\lim_{n\rightarrow\infty}\mathbb{P}(X_n=x)=0.$$ I tried some things with $P(X_n\leq x)=P(X_n<x)+P(X_n=x)$ and $P(X\leq x)=P(X<x)$, but don't see it...

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    $\begingroup$ You can use the fact that $F(x)=P[X\leq x]$ is nondecreasing in $x$ so it has at most a countable number of discontinuities. So for a fixed value $x$ at which $F$ is continuous, there are points $y_i<x$ that are arbitrarily close to $x$ and such that $F$ is also continuous at $y_i$ (why?) Can you answer the problem now? You can perhaps type your solution as an answer, which is standard practice when solving a question based on hints. $\endgroup$
    – Michael
    Commented Jan 17, 2018 at 22:10
  • $\begingroup$ Just try to follow @Michael 's hint (and btw Mau you need to tag him so that he notices). The hint pretty much says that there are points arbitralily close to $x$ where $F(x)$ is continuous. Can you conclude from this? Additional hint: What can you say now about $P[X_n < x + \varepsilon]$ $\endgroup$ Commented Jan 22, 2018 at 13:43
  • $\begingroup$ @HartoSaarinen Thank you. Well, taking a sequence $(x_k)$ of points of continuity converging to $x$ from below, I suppose we could argue $\lim_{n\rightarrow\infty}P(X_n=x)=P(X\leq x)-\lim_{n,k\rightarrow\infty}F_n(x_k)=P(X\leq x)-\lim_{k\rightarrow\infty}F(x_k)$, where the latter expression is zero if and only if F is continuous at $\lim_{n\rightarrow\infty} x_k=x$. If you mean that, I'm not sure how $P(X_n<x+\epsilon)$ is considered?! $\endgroup$
    – Mau314
    Commented Jan 22, 2018 at 14:12
  • $\begingroup$ @Michael Thank you for the hint. I posted a solution using the continuity property you're mentioning, but it seems to me you have an easier solution? $\endgroup$
    – Mau314
    Commented Jan 22, 2018 at 14:14
  • $\begingroup$ @Mau314 my hint was exactly what Michael wrote in his answer but instead of limit $i \to \infty$ I considered $\varepsilon \to 0$. $\endgroup$ Commented Jan 23, 2018 at 8:50

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I developed a solution that is perhaps more complicated than what Michael was thinking of, but I like how each ingredient is used very explicitly:

In the situation described above, we assume that $$\lim_{n\rightarrow\infty}\mathbb{P}(X_n=x)\neq 0,$$ which means that for some $\epsilon>0$ there is an infinite index set $I_\epsilon\subseteq\mathbb{N}$ such that $$\forall n\in I_\epsilon:~\mathbb{P}(X_n=x)\geq\epsilon,$$ so that in particular $$\forall n\in I_\epsilon~\forall x'<x:~F_n(x')\leq F_n(x)-\epsilon.$$ Furthermore, the fact that $\displaystyle\lim_{n\rightarrow\infty}F_n(x)=F(x)$ and the continuity of $F$ in $x$ tell us that for some $K\in\mathbb{N}$ and $\delta>0$ we have $$\forall k\geq K:~F_k(x)\leq F(x)+\frac{\epsilon}{3},~~\forall x'\in[x-\delta,x):~F(x)\leq F(x')+\frac{\epsilon}{3}.$$ Putting all this together, we see that for any element $m$ of the infinite index set $I_\epsilon\cap [K,\infty)$ and $x'\in[x-\delta,x)$ the following holds: $$F_m(x')\leq F_m(x)-\epsilon\leq F(x)-\frac{2\epsilon}{3}\leq F(x')-\frac{\epsilon}{3}.$$ Therefore, on the uncountable set $S=[x-\delta,x)$ we have $$\lim_{n\rightarrow\infty}F_n(\cdot)\neq F(\cdot).$$ However, by the definition of weak convergence, this entails that $F$ is discontinuous on $S$, which is impossible for any distribution function.

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  • $\begingroup$ This looks like a good alternative way. Below I fill in details on what I was thinking from my comment. $\endgroup$
    – Michael
    Commented Jan 22, 2018 at 17:45
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Mau answers it another way, here is what I had in mind from my comment: Let $x$ be a point of continuity of $F$. By my above comment, we know there is an infinite sequence $\{y_i\}_{i=1}^{\infty}$ such that $y_i<x$ for all $i \in \{1, 2, 3, ...\}$, $F$ is continuous at $y_i$ for all $i \in \{1, 2, 3, ...\}$, and $\lim_{i\rightarrow\infty} y_i = x$.

Now fix $n \in \{1, 2, 3, ...\}$ and $i \in \{1, 2, 3, ...\}$. By considering the interval $(y_i, x]$ we have $$P[X_n=x] \leq P[X_n\leq x] - P[X_n \leq y_i] $$ Taking a limit as $n\rightarrow\infty$ and using $F$ continuous at both $x$ and $y_i$: $$ \limsup_{n\rightarrow\infty} P[X_n=x] \leq F(x) - F(y_i) $$ Taking a limit as $i\rightarrow \infty$ and using the fact that $F$ is continuous at $x$ gives the result.

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  • $\begingroup$ Yes, thank you. That would be a more rigorous version of what I made of your hint in my comment above. $\endgroup$
    – Mau314
    Commented Jan 22, 2018 at 18:32

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