Two conflicting views on $1_{[n,n+1]}$ I need to justify whether $(f_{n})_{n}$ as a sequence of real-valued functions converges almost everywhere or not. Note that $f_{n}: \mathbb R \to \mathbb R$ where $f_{n}(x)=1_{[n,n+1]}(x)$. Now I have two views on this:
$1.$ $f_{n}$ converges pointwise to $0$ and thus $f_{n}(x) \xrightarrow{n \to \infty} 0$ for almost all $x$. This was my initial thought.
But then I started thinking about the fact that:
$2.$ for any $n \in \mathbb N$ we know that $\exists A_{n} \in \mathcal{B}(\mathbb R)$ so that $\lambda(A_{n})=1$ where $\lambda$ is the Lebesgue-Borel Measure on $\mathbb R$ and $\vert f(x)\vert \geq 1$ for all $x \in A_{n}$ and hence how can I say that it to zero almost everywhere, since there is always a part of the function of positive measure that is exactly a distance of $1$ away from the $0$ function. 
Any ideas that could illuminate this for me? 
 A: The sequence $(f_n)$ indeed converges pointwise to zero because  $f_n(x)\to 0$ for every $x$. That's all that pointwise convergence cares about: for each $x$, is it the case that $f_n(x)$ converges to $f(x)$? (Almost everywhere convergence asks: does the set of $x$ for which $f_n(x)$ fails to converge to $f(x)$ have zero measure? Clearly pointwise convergence implies almost everywhere convergence.)
What's interesting about this sequence is that the behavior of the $n$th function differs from the behavior of the limiting function $f(x):=0$. That is, the $n$th function has a bump, but the "final" function doesn't. In essence the bump "escapes to infinity". This phenomenon isn't limited to functions with unbounded domain. For example, the sequence of functions $g_n$ defined for $x\in[0,1]$ by $g_n(x):=nI_{(0,\frac1n)}(x)$ is a collection of rectangles, each with unit area. But you can check that $g_n$ converges to $0$ pointwise. (What happened to that unit area?!)
These examples are important because they illustrate that $\lim\int f_n$ isn't always equal to $\int\lim f_n$, i.e., you can't always interchange limits and integration. Integration theory, the next chapter after measure theory, gives a number of conditions under which that interchange is permissible.
A: We do have that $f_n(x) \to 0$ as $n\to \infty$ for all $x$, without the 'almost'.
This is more than enough, because convergence almost everywhere means that the sequence converges pointwise outside of a null set.
In this particular case, the sequence converges pointwise everywhere.
