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If the limit of a density function exists does it the follow that it is zero? To put it formally

$$\exists a \in \mathbb R \lim_{t \rightarrow \infty} f(t) = a \Rightarrow a= 0.$$

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1 Answer 1

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Yes.

Suppose the limit is anything else, so $\lim_{t \rightarrow \infty} f(t) = a \neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < \frac{a}{2}$. In particular, $f(t) > \frac{a}{2}$ in this reigon.

But then:

$$ \int_{\mathbf{R}} f(t) dt \geq \int_{N}^{\infty} f(t) dt \geq \int_{N}^{\infty} \frac{a}{2} dt = \infty $$

So $f$ cannot be a density function.

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