# Counterexample for $\limsup_n\int_0^1 f_n(x) \space dx \leq \int_0^1 \limsup_n f_n(x) \space dx$

Prove or Provide a Counterexample:

$$\limsup_n\int_0^1 f_n(x) \space dx \leq \int_0^1 \limsup_n f_n(x) \space dx$$

The solution says let $f_n(x)$ be the function with integral 1 that vanishes for $x \geq 1/n$.

Trying to understand this here. $\int_0^1 f_n(x) \space dx$ is 1 so the limit on the LHS is just 1. As for the RHS, if $\limsup_n f_n(x) = lim_{n\rightarrow \infty} \sup\{f_n(x), f_{n+1}(x), f_{n+2}(x), \dots \}$, my intuition says something like that is just the value infinity at the point $0$ only. Is this correct - and how can I formalize it?

• For $x>0$ the sequence $f_n(x)$ is eventually constant (for $n>1/x$) equal to zero. Therefore, not only $\limsup$, but $\lim_n f_n(x)=0$. For $x=0$, $\limsup_n f_n(0)=\infty$, but you don't even need to compute it since one point is irrelevant for the integral – uSir470888 Aug 12 '17 at 10:17
• what means "vanishes" here? That $\lim_n f_n(1/k)=0$ for any $k\in\Bbb N$? – Masacroso Aug 12 '17 at 10:19
• @Masacroso $f_n(x)=0$ for $x\geq 1/n$. – uSir470888 Aug 12 '17 at 10:23
• @user470888 I see... the word "vanish" reminds me a process (in contraposition to a state) that make something zero, this is why it confuses me a bit. It is more clear just say that $f_n(x)=0$ for $x\ge 1/n$. – Masacroso Aug 12 '17 at 10:26

Consider $f_n(x)=\frac{1}{n}\chi_{[0,n]}$. Then $\limsup(f_n)=0$ since as n goes to infinity $f_n$ tends to the function identically zero on the interval $[0,\infty]$ (here if you draw the graph of the function it becomes clearer). On the other hand the integral of $f_n$ is the area below the graph, hence $\int f_n(x)dx=\frac{1}{n}\cdot n=1$, obtaining the counterexample.
If for each $f_n$ we have $\int \limits_{0}^1 f_n(x)dx=1$ and $f_n(x)=0$ for $x\geq \frac{1}{n}$ then clearly we have $\limsup\limits_n f_n(x)=0$ for all $x>0$. It follows that:
$1=\limsup_n\int_0^1 f_n(x) \space dx > \int_0^1 \limsup_n f_n(x) \space dx=0$.
For an example of such an $x_n$ consider the triangle function:
$f_n(x)=2(n-nx)$ if $x\leq \frac{1}{n}$ and $f_n(x)=0$ if $x\geq \frac{1}{n}$