Complete convergence not happening but convergence in probability occurs So today I created a counterexample to "Convergence in Probability implies Almost Sure Convergence". I considered a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ and $P(X_n=0)=1-\dfrac{1}{n}$.
As is evident, this sequence $X_n$ converges in probability to $0$ but does not converge "completely", so by using Second Borel-Cantelli Lemma and converse of Complete Convergence, $X_n$ does not converge almost surely.
I feel my example is correct and things are well-defined. However, when I told my professor of this counterexample, he said that I need to specify the probability space i.e. $\{\Omega,A,P\}$ where $\Omega$ is the sample space, $A$ is the usual collection of events and $P$ is the probability function.
He further added that I must show him, by constructing such a probability space, that $X_i$ independent of $X_j$ whenever $i\neq j$ happens, i.e. I am not saying anything fancy or unachievable.
Now, I have never worked with probability spaces before. I don't know how to construct one. Please tell me what more I need to tell my professor to convince him that my example is a valid counterexample. I will be able to understand what you mean if you completely specify the probability space. Thank you.
 A: Your professor is right.  The point is that the statement: "a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ and $P(X_n=0)=1-\dfrac{1}{n}$" is not enough to define a sequence of random variables.  
You need to PROVE that there is "a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ and $P(X_n=0)=1-\dfrac{1}{n}$". Or even better: explictly present the sequence.
Consider $\{[0,1],A,P\}$ where the closed interval $[0,1]$ is your sample space, $A$ the Borel $\sigma$-algebra is the set of events, and $P$ is the usual length measure (Lebesgue measure).  
The random variables $X_n$ are functions (actually measurable functions) defined in $[0,1]$ with values in $\mathbb{R}$, for instance.  
Can you describe the random variables in your example?  
Remark: It is true that convergence in probability does not imply almost sure convergence, and there are known examples showing it. But, I understand that the question here is about the example you are proposing. 
A: Your  example should be $P(X_n=1)=1/n $ and $P(X_n=0)=1-1/n$.
To show that $X_n$ does not converge almost surely. Here you can do 
Let $A_n(\epsilon)=\{ |X_n|\geq\epsilon\}$
and 
let $B_m(\epsilon)=\cup_{n\geq m} A_{n}(\epsilon)$
Hint:


*

*Show that $X_n\to 0$ a.s iff $P(B_m(\epsilon))\to 0$ as $m\to \infty$

*From your example, show that $P(B_m(\epsilon))=1 $ for all $m$.


You can calculate $P(B_m(\epsilon))$ as follow
$P(B_m(\epsilon))=1-P(X_n=0 \ \text{for all} \ m\leq n)$
$P(B_m(\epsilon))=1-(1-\frac{1}{m})(1-\frac{1}{m+1})\ldots$
$P(B_m(\epsilon))=1-\lim_{k\to\infty}(\frac{m-1}{m}\frac{m}{m+1}\ldots \frac{K}
{K+1})$
$P(B_m(\epsilon))=1-\lim_{k\to\infty}(\frac{m-1}{K+1})=1$
A: Landon, here is an example that matches what you are looking for and will also satisfy what your professor is asking for. 
Consider $\{[0,1],A,\lambda\}$ where the closed interval $[0,1]$ is your sample space, $A$ the Borel $\sigma$-algebra is the set of events, and $\lambda$ is the usual length measure (Lebesgue measure).  
Let $\{[0,1]^{\mathbb{N}}, B,P\}$ be countable infinite product space where each factor is $\{[0,1],A,\lambda\}$. 
For the construction and existence of such product space, see Halmos $\S$38. 
Each $\textbf{x}\in [0,1]^{\mathbb{N}}$ is a sequence $(x_0, x_1, x_2, ...)$, where, for each $i\in \mathbb{N}$, $x_i \in [0,1]$.
For  each $i\in \mathbb{N}$, there is a function $\pi_i$ from $[0,1]^{\mathbb{N}}$ onto $[0,1]$ such that $\pi_i(\textbf{x})=x_i$.
Now we can define the random variables. 
For  each $n\in \mathbb{N}$, $X_n$ is a random variable (a function) from $[0,1]^{\mathbb{N}}$ to $\mathbb{R}$ defined by
$X_n(\textbf{x})=n$ if $\pi_n(\textbf{x})\in[0,\frac{1}{n}]$
$X_n(\textbf{x})=0$ if $\pi_n(\textbf{x})\notin[0,\frac{1}{n}]$
It easy to see that each $X_n$ is actually a random variable (it means: it is a measurable function) and that the set $\{X_n\,|\,n\in \mathbf{N}\}$ of random variables are independent. 
So we have explicitly presented a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ and $P(X_n=0)=1-\dfrac{1}{n}$. 
It is easy to seem that this sequence converge in probability to 0, and NOW, using your argument, we can also see that it does NOT converge almost surely to 0.  
Remark: The example above is a nice example of sequence of random variables converging in probability but not converging almost surely. The only drawback that I see is that it is far more complex than the usual example in literature (for instance: see Halmos $\S$22 exercise 6).
On the other hand, in favor of the example above, we have that it shows that even if we assume the sequence of random variables to be independent, the sequence may converge in probability and do not converge almost surely.   
