A problem about stochastic convergence (I think) I am trying to prove the convergence of the function $f_n = I_{[n,n+1]}$ to $f=0$, but first of all I don't in which way it converges, either in $\mathcal{L}_p$-measure or stochastically, or maybe some other form of convergence often used in measure-theory.
For now I'm assuming it's stochastic convergence, as in the following:
$$ \text{lim}_{n \rightarrow \infty} \, \mu(\{x \in \Re: |f_n(x)-f(x)| \geq \alpha\}\cap A )=0$$
must hold for all $\alpha \in \Re_{>0}$ and all $A \in \mathcal{B}(\Re)$  of finite measure.
I know it must be true since there is no finite $A$ for which this holds. Could someone give me a hint how to start off this proof?
 A: The sequence $\{f_n\}$ doesn't converge in $\mathcal L^p$ norm, since for all $n$ $$\lVert f_{n+1}-f_n\rVert_{L^p}^p=\int_{\mathbb R}|\mathbf 1_{[n+1,n+2]}-\mathbf 1_{[n,n+1]}|^p =\int_{[n,n+2]}1d\mu =2.$$
This sequence cannot converge in measure since $\mu(\{|f_{n+1}-f_n|\geq \frac 12\})\geq \mu([n,n+1))=1$, but converges pointwise to $0$. 
It also converges stochastically to $0$, since if $\alpha> 1$, we have $\{x\in\mathbb R\mid |f_n|\geq \alpha\}=\emptyset$. For $\alpha\leq 1$, and $A\in\mathcal B(\mathbb R)$ with $\mu(A)<\infty$, use the fact that 
$$\mu(A)\geq \mu(A\cap \mathbb R_+)=\sum_{n=0}^{+\infty}\mu(A\cap[n,n+1]).$$
A: For those interested, this is my complete proof:
For $\alpha > 1$ the limit is clearly equal to $0$. For $\alpha \leq 1$ we need to prove that 
$$ \text{lim}_{n \rightarrow \infty} \mu([n,n+1] \cap A) = 0.$$
Let us assume that there is an $A \in \mathcal{B}(\mathbb{R})$ with finite measure, such that the above expression is not equal to $0$, but some $c \in \mathbb{R}$. If we take $\varepsilon \in \mathbb{R}_{>0}$, then there exists an $N \in \mathbb{N}$ such that for all $n>N$ we have $|\mu([n,n+1] \cap A) - c| < \varepsilon$. We can then make the following estimation for all $n>N$:
$$\mu([n,n+1] \cap A) > c - \varepsilon.$$
And since $\mu(A) \geq \mu(A \cap \mathbb{R}^+) \geq \sum_{n=N}^\infty \, \mu([n,n+1] \cap A) = \infty$ we have a contradiction, since we assumed $A$ had finite measure. Therefore the limit is always equal to $0$ for all finite $A \in \mathcal{B}(\mathbb{R})$ and $f_n$ converges to $f$.
