# To find suitable example regarding Fatou's lemma

I am asked the following question:

By suitable example, show that strict inequality holds in Fatou's lemma

I know that Fatou's lemma states that

Suppose $f_n$ is a sequence of measurable functions with $f_n$ non-negative. If $f_n$ coverges to $f$ for a.e $x$ then $\int f=\liminf\int f_n$.

I need to construct an example to show that strict inequality holds in lemma. Please help. I don't have idea to start.

Actually, Fatou's lemma asserts that if $\{f_n\}$ is non-negative, then $$\int_X\liminf_{n\to\infty}f_n\leq \liminf_{n\to\infty}\int_Xf_n$$

To show that strict inequality can hold, consider $f_n=1_{[n,n+1]}$ on $\mathbb{R}$ with Lebesgue measure.

• Please tell me clearly how you took fn function? fn = 1[n,n+1] on R?
– K.S
Nov 12 '16 at 6:55
• $1_{[n,n+1]}(x)=1$ if $n\leq x\leq n+1$, and is zero otherwise. Therefore $f_n(x)\to 0$ for all $x$, but $\int f_n=1$ for each $n$. Nov 12 '16 at 6:56
• You forgot something while writing the comment as I see the comment in this fashion:-. $1_{[n,n+1]}(x)=1$ if $n\leq x\leq n+1$, and is zero otherwise. Therefore $f_n(x)\to 0$ for all $x$, but $\int f_n=1$ for each $n$.
– K.S
Nov 12 '16 at 7:12
• I don't think I forgot anything. Which part is unclear? Nov 12 '16 at 7:16
• Use of mathjax is not properly done I guess
– K.S
Nov 12 '16 at 7:19

The intuition I hold behind Fatou's lemma is that if you have a pointwise limit of functions, then there can be no new mass created. If we think of the value of the integral as the mass, you can't have a sequence of functions and then end up with some new mass. That isn't to say you can't lose mass. The mass could end up scooting down the real line and disappearing at infinity.

With this intuition, the counterexample is easy: take a bump function, as carmichael561 said, and its translates. The mass for each function remains the same, but in the limit it ends up running off to infinity.