when does a continuous PDF NOT have a limit at infinity I have to find out if it is correct, that if a continuous random variable has an expected value and it's PDF is continuous, then
$$\lim_{t \to +\infty}f_{\xi}(t)= 0$$
I know how to prove that the limit is equal to 0 if we presume it exists. But I'm struggling to understand what happens if it doesn't or how to prove that the limit exists.. I would appreciate any help.
Edit: Thanks for all the answers and comments. They helped me realize what I don't understand: what could be an example of a continuous PDF which has a mean and which limit is not existent?
Edit 2: I've found an example of a continuous function, which integral is 1, but I'm not sure if it has an expected value. Is it possible to calculate it for such a function?
Here's
its graph
 A: Like one of the comments say, there could be a counter example. In this answer i am trying to make progress in the positive direction.
$\int_{0}^{\infty} t f_{\zeta} (t) dt = \text{constant} < \infty \implies \lim_{a \rightarrow N}\int_{a}^{\infty} t f_{\zeta} (t) dt \rightarrow 0$, Let $a = N$ then $s_N = \int_{N}^{\infty} t f_{\zeta} (t) dt $ and $\lim_{N \rightarrow \infty} s_N = 0$. Now since $f_{\zeta} (t) \geq 0$, $\forall t$ => $s_N$ is monotonically decreasing and converges to $0$, in particular $|s_n| < \epsilon_N$, $\forall n \geq N$. Define $E_{mN} = \{x: x \geq N, |f_{\zeta}(x)| > \frac{1}{m}\}$. Let $\mu$ be lebesgue measure. $$\mu(E_{mN}) \frac{N}{m}   < |\int_{E_{mN}} t f_{\zeta} (t) dt | \leq |\int_{N}^{\infty} t f_{\zeta} (t) dt | < \epsilon_N.$$ Hence $$\mu(E_{NN}) < \epsilon_N.$$ Hence $$0\leq \lim_{N \rightarrow \infty} \mu(E_{NN}) \leq  \epsilon_N \rightarrow 0.$$ Hence $$\lim_{N \rightarrow \infty} \mu(E_{NN}) = 0.$$ This seems to be the best i can prove. To prove what you want, we need: $$\lim_{N \rightarrow \infty} \mu(\cup_{m \geq N} E_{mN}) = 0.$$ Thanks.
EDIT:
One of the comments above gives a counter example. So the above seems to be the be best we can prove as a positive result.
