Question: Does $\{\chi\_{[n, n+1]}\}$ converge to $0$ in measure?

I want to determine if it does or not using the definition of convergence in (Lebesgue) measure, which is:

> 
> $f\_{n} \to f$ in measure as $n \to \infty$ if for any $\epsilon > 0$,
> $$m(\{x \in \mathbb{R} \mid |f\_{n}(x)-f(x)|>\epsilon\}) \to 0
>  \;\text{as}\; n \to \infty$$
> 
> 
> 

To me it seems that the answer should be yes because given any $x \in \mathbb{R}$ and $\epsilon >0$, we can find some $N \in \mathbb{N}$ such that $x \notin [n, n+1]$ for all $n \ge N$. Then the measure of the set $\{x \in \mathbb{R} \mid |\chi\_{[n,n+1]}|>\epsilon\}$ is zero as $n \to \infty$. Am I right?

Answer: 
If $0 < \epsilon < 1$, then the set $\{x \in \mathbb{R} \mid |f\_n(x) - f(x)| > \epsilon\}$ is simply $[n, n+1]$, since $|f\_n(x)-f(x)| = \chi\_{[n, n+1]}(x)$. The question boils down to whether $m([n, n+1]) \to 0$ holds (which it does not).
Under the definition given in the OP $\mathbb{1}\_{[n,n+1]}$ does not converge to $0$ in probability (relative to the Lebesgue measure) since for any $0<\varepsilon<1$, $m(\mathbb{1}\_{[n,n+1]}>\varepsilon)=m([n,n+1])=1$.

The definition if convergence in measure given in the OP is not useful when the underlaying measure space is infite, that is when $(X,\mathscr{B},\mu)$ is a measure space and $\mu(X)=\infty$. A more useful notion of convergence in measure is the following:

Suppose $(X,\mathcal{B},\mu)$ is a measure space. A sequence $f\_n$ of measurable functions converges in measure to another function $f$ if for any set $A\in\mathcal{B}$ of finite measure and any $\varepsilon>0$
$$\mu(A\cap\{|f\_n-f|>\varepsilon)\xrightarrow{n\rightarrow\infty}0$$

**Remarks:**

* Under this *local* version of convergence in measure the sequence in the OP does converge to $0$ due to the regularity of the Lebesgue measure.
* The *local* version of convergence in measure given here coincides with the usual one given in probability courses.