0

Is it possible to create a positive continuous function with non-zero limits in $+\infty$ and $-\infty$ whose integral over $\mathbb{R}$ is $1$?

I am studying the probability density functions and one of the conditions mentioned are that limits of the function in $+\infty$ and $-\infty$ must be $0$. Can't wrap my head around that.

Bobson Dugnutt
  • 10,672
waterlemon
  • 189
  • Consider the related question of when $\lim\limits_{n\to\infty}\sum\limits_{i=0}^n a_i$ converges. Let $A_n=\sum\limits_{i=0}^na_i$ be the sequence of partial sums. If it were to converge, then there should be some $N$ for which $m,n>N$ you have $|A_m-A_n|<\epsilon$ for each $\epsilon>0$. But that implies $|A_{n+1}-A_n|=|a_{n+1}|<\epsilon$, so the terms must approach zero. The proof is similar for integrals instead of sums, depending on whether you use riemann or lebesgue integration. – JMoravitz Mar 12 '16 at 14:40
  • you could get a function that has no limits – Mirko Mar 12 '16 at 18:21

2 Answers2

5

If one limit or the other at infinity exists, then it must be zero. Otherwise you can choose large $M$ so that $f>\varepsilon$ for $x>M$ (or $x<-M$, depending), in which case $\int_M^\infty f = +\infty$. Then the overall integral can't converge.

However, you can have a probability density with no limit at infinity. For example, if $f(x)=\begin{cases} 4x & 0 \leq x \leq 1/2 \\ 2-4(x-1/2) & 1/2 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$

then $g(x)=\sum_{n=1}^\infty f(2^n(x-n))$ is a probability density (and is a continuous function). Its graph looks like infinitely many triangles of height $2$ and width $2^{-n}$, so they have area $2^{-n}$ for $n=1,2,\dots$. It has no limit at infinity, because $g(n+2^{-n-1})=2$ for every natural number $n$, but $g(n)=0$ for all natural numbers $n$. (Also, this function is continuous and nonnegative, so there is no issue of Riemann vs. Lebesgue.)

It takes some work, but you can even make such a function infinitely differentiable. Notably it cannot be made uniformly continuous.

Ian
  • 101,645
0

It is, indeed, not possible and not very difficult to see if you interpret the integral as the area under the curve. For a formal argument, consider $f: \mathbb R \to \mathbb R_+$ such that $f$ has a nonzero limit at $+\infty$, say $2l >0$. Then, there is an $A > 0$ such that $f(x) > l$ for each $ x > A$. Now, for each $ x > A$, we can write $\int_0^x f \ge \int_A^x f \ge lx$. Consequently, $f$ is not integrable.

Bobson Dugnutt
  • 10,672
KGD
  • 163