**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.