Cdf of a discrete random variable and convergence of distributions

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Artusartos
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In the page that I attached, it says "...while at the continuity points x of [itex]F_x[/itex] (i.e., [itex]x \not= 0[/itex]), [itex]lim F_{X_n}(x) = F_X(x)[/itex]." But we know that the graph of [itex]F_X(x)[/itex] is a straight line y=0, with only x=0 at y=1, right? But then all the points to the right of zero should not be equal to the limit of [itex]F_{X_n}(x)[/itex], right? Because [itex]F_X(x)[/itex] is always zero at those points, but [itex]F_X(x)[/itex] is 1? So how do I make sense of that?

Thanks in advance
 
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Artusartos said:
But we know that the graph of [itex]F_X(x)[/itex] is a straight line y=0, with only x=0 at y=1, right?

No, I think [itex]F_X(x)[/itex] is the cumulative distribution, not a density function.
 
Stephen Tashi said:
No, I think [itex]F_X(x)[/itex] is the cumulative distribution, not a density function.

Oh, ok...

But it's still confusing. What if n=4 (for example)? Then [tex]F_{X_n} = 1[/tex] if [tex]x \geq 1/4[/tex], and [tex]F_{X_n}=0[/tex], when [tex]x < 1/4[/tex], right? So for any x between 0 and 1/4, the limit at those points is 0, but the limit of [tex]F_X[/tex] at those points is 1...so the limits are not equal, are they?
 
Artusartos said:
So for any x between 0 and 1/4, the limit at those points is 0,

What limit are you talking about? Something like [itex]lim_{x \rightarrow 1/8} F_{X_4}(x)[/itex] ? I see nothing in the discussion in the book that dealt with that sort of limit. The limits under consideration involve [itex]n \rightarrow \infty[/itex].