Let $\nu$ be a $\sigma$-finite measure on $\mathbb R$ with $$\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag1$$ and $\nu(\{0\})=0$, $$I_k:=\left(-\frac1k,-\frac1{k+1}\right]\cup\left[\frac1k,\frac1{k+1}\right)$$ and $$\nu_k(B):=\nu(B\cap I_k)\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $k\in\mathbb N$ and $(X_k)_{k\in\mathbb N}$ be a real-valued independent process on some probability space with $$\operatorname E[X_k]=\int\nu_k({\rm d}x)x=\int_{I_k}\nu({\rm d}x)\tag2$$ and $$\operatorname{Var}[X_k]=\int\nu_k({\rm d}x)x^2=\int_{I_k}\nu({\rm d}x^2)\tag3$$ for all $k\in\mathbb N$ and $$M_n:=\sum_{k=1}^n\left(X_k-\operatorname E[X_k]\right)\;\;\;\text{for }n\in\mathbb N.$$
It's easy to see that $M$ is a martingale with $$\operatorname E\left[M_n^2\right]=\sum_{k=1}^n\operatorname{Var}[X_k]\;\;\;\text{for all }n\in\mathbb N\tag5$$ and hence $$\sup_{n\in\mathbb N}\operatorname E\left[M_n^2\right]=\sum_{k\in\mathbb N}\operatorname{Var}[X_k]=\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag6.$$
Why can we conclude that $(M_n)_{n\in\mathbb N}$ is almost surely convergent as $n\to\infty$?
The martingale convergence theorem, for $L^2$-bounded martingales.