Why $f_n(x)=\chi_{[n,n+1]}$ not converges almost uniformly in $\mathbb{R}$

Question

Let $(\mathbb{R},\mathbb{B}, \lambda )$ the real line with Lebesgue measure on the Borel Subsets of $\mathbb{R}$.

why the sequence $f_n(x)=\chi_{[n,n+1]}$ not converges almost uniformly in $\mathbb{R}$?

I was able to show that the sequence converges everywhere to 0, But I can not see how to use the definition of being almost uniformly convergent to come to a contradiction or how to use directly. Can anyone help?

Answer 1

Suppose $\lambda (E) <1$ and $f_n \to 0$ uniformly on $E^{c}$. Then $f_n(x) <1$ for all $x \in E^{c}$ for $n$ sufficiently large. In particular , $f_n(x) <1$ for all $x \in E^{c}$ for some integer $n$. But then $E^{c}$ cannot contain any point of $[n,n+1]$. In other words, $[n,n+1] \subset E$ which implies $\lambda (E) \geq 1$, a contradiction.

Answer 2

Note that $\lVert f_n\rVert_{\infty}=1$ (norm is the sup-norm) for all $n$. Hence it can't converge almost uniformly to zero.