# A sequence $f_k:\Omega\rightarrow \mathbb R$ such that $\int f_k=0 \quad \forall k\in \mathbb N$ and $\lim\limits_{k\to\infty} f_k \equiv1$.

Find a sequence of Lebesgue Integrable functions $f_k:\Omega\rightarrow \mathbb R$ such that
$\int f_k=0 \quad \forall k\in \mathbb N$ and $\lim\limits_{k\to\infty} f_k \equiv1$.

My guess is $$f_k(x) = sign(x-k)$$ Then integral is $0$ by symmetry and $f_k$ converges to 1. Is it right?
Can you give more examples?

• Lebesgue measurable? Then what is $\Omega$ here? – drhab Feb 15 '18 at 16:21

## 2 Answers

If the functions are supposed to be Lebesgue integrable over $\mathbb R$, then your example, whose absolute value is $1$ everywhere, is not Lebesgue integrable.

An example that illustrates the principle would be $$f_n(x) = \begin{cases} 1 & x \in [-n, n] \\ -2n & x \in [-(n+1),-n) \\ 0 & \text{else}. \end{cases}$$

• Is $sign$ not Lebesgue integrable because $\int_\mathbb R |sign()| \ dx = \infty$? – user3342072 Feb 15 '18 at 17:11
• Yes, precisely. – AlgebraicsAnonymous Feb 15 '18 at 17:21

If $\Omega=[0,1]$, then you can set $$f_k(x) = \begin{cases} 1 & x \in [\frac{1}{k}, 1] \\ 1-k & x \in (0,\frac{1}{k}) \\ 1 & x = 0 \end{cases}$$