**Question summary.** Does the Kolmogoroff condition $\sum\_{n=1}^\infty\frac{\mathbb V Y\_n}{n^2} < \infty$ hold for truncated random variables $Y\_n := X\_n \cdot 1\_{\{X\_n \le n\}}$ (see below for a more rigid definition)?


**General Definitions.** Let $(\Omega, \mathcal A, \mathbb P)$ be a probability measure space, let $\mathbb E$ denote the expected value and $\mathbb V$ the variance. Let $1\_\text{set}$ be the characteristic function of $\text{set}$.


**Definition.** Let $(X\_n)\_{n\in\mathbb N}$ be a sequence of non-negative, pairwise independent real random variables with finite expected values such that $\eta :=\displaystyle \lim\_{n \to \infty} \mathbb E X\_n$ exists. (Then clearly $\sup\_n \mathbb E X\_n < \infty$.) Let $Y\_n:= X\_n \cdot 1\_{\{X\_n \le n\}}$. 


I want to generalize the following result:


**Theorem 1.** *If the $X\_n$ are identically distributed, then $\displaystyle\sum\_{n=1}^\infty\frac{\mathbb V Y\_n}{n^2} < \infty$*.


To a Theorem where the $X\_n$ need not be identically distributed:


**Statement 2.** (is this true?) *$\ \displaystyle\sum\_{n=1}^\infty\frac{\mathbb V Y\_n}{n^2} < \infty$ without further restrictions on the $X\_n$.*


**Proof of Theorem 1.** Consider any random variable $M$ with the same distribution as all the $X\_n$. Then
\begin{equation}
 \begin{split}
 \sum\_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb V Y\_j 
 & \le \sum\_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1\_{\{M \le j \}})
 = \lim\_{N\to\infty} \sum\_{j=1}^N \left(\frac{1}{j^2} \sum\_{k=0}^{j-1} \mathbb E(M^2 \cdot g\_k)\right)\\
 & = \lim\_{N\to\infty} \sum\_{k=0}^{N-1} \left(\mathbb E (M^2 \cdot g\_k) \sum\_{j=k+1}^N \frac{1}{j^2}\right)\\
 & \le c + \lim\_{N\to\infty} \sum\_{k=1}^{N-1} \left(\mathbb E(M^2 \cdot g\_k) \sum\_{j=k+1}^N \frac{1}{(j-1) j}\right)\\
 & \le c + \lim\_{N\to\infty} \sum\_{k=1}^{N-1} \mathbb E(M^2 \cdot g\_k) \cdot \frac{1}{k}
 = c + \sum\_{k=1}^{\infty} \frac{1}{k} \cdot \int\_{\{k < M \le k + 1 \}} M^2 \, \mathrm d \mathbb P\\
 & \le c + \sum\_{k=1}^\infty \frac{k+1}{k} \cdot \mathbb E{(M \cdot g\_k)}
 \le c + 2 \cdot \mathbb E M < \infty \qquad \square
 \end{split}
 \end{equation}


I initially thought that I could use the same proof for Statement 2 by considering $M := \sup\_n X\_n$ (by Beppo Levi we would have $\mathbb E M < \infty$.) The exact proof works except for the very first inequality (since $\frac{1}{j^2} \cdot \mathbb V Y\_j \le \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1\_{\{X\_j\le j \}})$ is still true but $\frac{1}{j^2} \cdot \mathbb V Y\_j \le \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1\_{\{M\le j \}})$ is wrong in general.)


**Proving the following Lemma would be enough:**


**Lemma.** *If $\sum\_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1\_{\{M\le j \}})<\infty$ then $\sum\_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb V Y\_j < \infty$*


*Some ideas for the proof of this Lemma.* We have 


\begin{equation}
\begin{split}
 \mathbb V Y\_j -\mathbb E(M^2 1\_{\{M\le j \}}) &= \mathbb E(X\_j^2 1\_{\{X\_j\le j\}})-\mathbb E(X\_j 1\_{\{X\_j\le j\}})^2 -\mathbb E(M^2 1\_{\{M\le j \}}) \\
&\overset{\text{(\*)}}\le \int\_{\{X\_j\le j\}\setminus\{M\le j\}} X\_j^2 \,\mathrm d \mathbb P \le j^2 \cdot \mathbb P(\{X\_j\le j\}\setminus\{M\le j\})\\
&\le j^2 \cdot\mathbb P\{M>j\} \overset{\text{Markow}}\le const \cdot j
\end{split}
\end{equation}
(\*): I feel like I am loosing something in this inequality


However, this gives an additional $1/j$ term in the sum and is thus not enough to prove the lemma.



Kolmogoroff condition for truncated random variables
Welcome to MathOverflow! However, your conjecture is false. Indeed, let $P(X\_n=n)=1/n=1-P(X\_n=0)$. Then for all $n$ we have $EX\_n=1$, $Y\_n=X\_n$, $Var\, Y\_n=Var\,X\_n=n-1$. So, $\sum\_n Var\,Y\_n/n^2=\infty$. 




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Additional note: Your statement that $EM<\infty$ does not follow from the Beppo Levi theorem, and it is actually false in general. Indeed, in the above example, by the [second Borel–Cantelli lemma](https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma#Converse_result), the events $\{X\_n=n\}$ occur almost surely (a.s.) infinitely often, and hence $M=\infty$ a.s. 


