Limit of the function vs. limit of its average

Question

Is it true that if $(b-a)$ is fixed, then $\lim \limits_{a,b \to \infty}\frac{1}{b-a}\int_a^b f(t)dt=\lim \limits_{t \to \infty}f(t)$? The function $f(t)$ isn't too nice, it's stochastic, involves jumps etc., so mean value theorem wouldn't work.
Intuitively, it should be true. If it isn't in general, are there any conditions under which it would be true?

Answer 1

If $b-a$ is fixed, then it's fixed to some constant $c$

The limit then becomes

$$\lim_{a\to\infty}\frac{1}{c}\int_a^{a+c} f(t)dt$$

Now, if $\lim_{t\to\infty} f(t)$ exists and is equal to $L$, then we know that for each $\epsilon>0$, there exists some $M>0$ such that if $x>M$, then $|f(x)-L|<\epsilon$.

Therefore, if $a>M$, you know that $$\int_a^{a+c} f(t)dt < \int_a^{a+c} (L+\epsilon)dt = c(L+\epsilon)$$

and $$\int_a^{a+c} f(t)dt > \int_a^{a+c} (L-\epsilon)dt = c(L-\epsilon)$$

which should be enough to prove your statement.

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However, if the limit of $f$ does not exist, then the limit of the integral can still exist, but then the equality does not hold anymore (since one side of the equality is undefined)

Answer 2

Define $f: \mathbb R \to \mathbb R$ by

$f(x)=1$ , if $x \notin \mathbb Z$ and $f(x)=0$ if $x \in \mathbb Z$ .

If $b>a$, then $\frac{1}{b-a}\int_a^b f(t)dt=1$, but the limit $\lim_{x \to \infty}f(x)$ does not exist.