Show that this process is not a martingale I have to show that the following process $(X_t)_{t\in [0,\infty)}$ is no martingale.
Let $Y_n$ be a sequence of independent random variables with
$$P(Y_n=n)=\frac{1}{2n^2},\quad P(Y_n=-n)=\frac{1}{2n^2},\quad P(Y_n=0)=1-\frac{1}{n^2}.$$
We put
$$X_t=\sum_{\left\{n\,;\, 1-t\leq \frac 1{n}\right\}} Y_n\quad\text{for all } t\geq 0.$$
An ansatz would be to assume that is a martingale and then, by the fundamental theorem of local martingales, you can decompose $X$ as
$$X=A+B$$
where $A$ is a local martingale with bounded jumps and $B$ is a process of locally integrable variation.
But we have
$$\sum_{\{s\leq t\,:\,\Delta X_s\geq 1\}}|\Delta X_s|=\infty,$$
which contradicts $B$ having integrable variation. But it does not necessarily contradict $B$ having locally integrable variation.
Any help is appreciated!
 A: Set
$$S_k := \sum_{n=1}^k Y_n.$$
Since
$$\sum_{n \geq 1} \mathbb{P}(Y_n \neq 0) = \sum_{n \geq 1} \frac{1}{n^2} < \infty,$$
it follows from the Borel Cantelli lemma that for almost all $\omega \in \Omega$ we have $Y_n(\omega)=0$ for all $n \geq N(\omega)$ sufficiently large. This implies, in particular, that
$$S_{\infty} := \lim_{k \to \infty} S_k = \sum_{n \geq 1} Y_n$$
exists almost surely (as a pointwise limit).

Claim: $S_{\infty}$ is not integrable. In particular, $(S_n)_{n \in \mathbb{N} \cup \{\infty\}}$ is not a martingale.

Proof: For $n \in \mathbb{N}$ define $$A_n := \{Y_n = n\} \cap \bigcap_{k \neq n} \{Y_k=0\}.$$ Since $$a:=\prod_{k \geq 1} \left(1- \frac{1}{k^2} \right) > 0,$$ it follows from the independence of the random variables $(Y_k)_{k \in \mathbb{N}}$ that $$ \mathbb{P}(A_n) = \mathbb{P}(Y_n=n) \prod_{k \neq n} \mathbb{P}(Y_k=0) \geq \mathbb{P}(Y_n=n) \prod_{k \geq 1} \mathbb{P}(Y_k=0)= a \frac{1}{2n^2}$$ for all $n \in \mathbb{N}$. As the sets $A_n$ are pairwise disjoint, we find from the monotone converence theorem that $$\begin{align*} \mathbb{E}(|S_{\infty}|) \geq \mathbb{E} \left(|S_{\infty}| \sum_{n \in \mathbb{N}} 1_{A_n} \right) &= \sum_{n \in \mathbb{N}} \mathbb{E}(|S_{\infty}| 1_{A_n}) \\ &= \sum_{n \in \mathbb{N}} n \mathbb{P}(A_n) \\ &\geq a \sum_{n \in \mathbb{N}} n \frac{1}{2n^2} = \infty \end{align*}$$ which proves the assertion.
Finally, we note that by the very definition of $(X_t)_{t \geq 0}$, it holds that
$$X_1 = \sum_{n \in \mathbb{N}} Y_n = S_{\infty};$$
since we have shown that $X_1$ fails to be integrable, the process $(X_t)_{t \geq 0}$ is not a martingale.
