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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?

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Haha, I´ve been looking at this problem for a little too long. It´s pretty clear now that there is no $A \in \mathcal{B}(\Re)$ such that $\mu([n,n+1]\cap A) \neq 0$, since $\mathcal{B}(\Re)$ only has "local sets". I'll work out the last details now, thanks! –  BallzofFury Dec 19 '11 at 13:20
    
Ballz: where did you get the strange notion that Borel sets were only "local" (whatever that means)? –  Did Dec 19 '11 at 13:31
    
Good point, but since the measure of each individual $A_n$ tends to $0$, the limit of the measure of the intersection still goes to 0 as $n$ gets larger. I'll try to show that if there is an $A$ for which $\mu([n,n+1] \cap A) \neq 0$, it must have infinite measure. –  BallzofFury Dec 19 '11 at 13:39
    
@Didier: Still getting farmiliar with Borel sets, Davide's example shows your point. –  BallzofFury Dec 19 '11 at 13:42
    
Ballz: That was not my point nor my question. –  Did Dec 19 '11 at 13:48
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2 Answers

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 $$\mu(A)\geq \mu(A\cap \mathbb R_+)=\sum_{n=0}^{+\infty}\mu(A\cap[n,n+1]).$$

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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 $f$.

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