Usbrth

Let $p in (0,1)$. If $P(|a|>k)leq p^k$ then $$lim_kto inftyint_a^2dmathbbP = 0?$$

is this affirmation true? It seems to be, but as I cannot use monotone convergence theorem, nor dominated convergence theorem, as the measurable function a might not be bounded, I'm lost on how to show it. Is the exponential bound on the probability enough to ensure the limit is $0$?

Sorry for not giving details of any attemps but as I cannot bound the integrand by $k$, nor anything related, I'm unsure how to proceed.

Thanks in advance.

Yes, this is true. Break the sum up as:
beginalign*
int_a^2 d mathbbP&=sum_n=k^infty int_n< a^2 d mathbbP\
&leq sum_n=k^infty (n+1)^2 mathbbP(n<|a|leq n+1)\
&leq sum_n=k^infty (n+1)^2 p^n.
endalign*

Now the sum above is convergent for $p in (0,1)$, so in particular the tail sums have to be going to zero.

A variation on kccu's solution uses the trick that for any random variable $a$ and any $q > 0$, you can write
$$int |a|^q,dmathbbP = q int_0^infty x^q-1 mathbbP(|a| > x),dx.$$
To prove this, write $mathbb P(|a| > x) = int 1_ > x,dmathbbP$ and interchange the integrals (which is justified since everything is nonnegative). Then note that $int_0^infty q x^q-1 1_ > x,dx = int_0^a q x^q-1,dx = |a|^q$ by the fundamental theorem of calculus.

With $q=2$ this reads
$$int a^2,dmathbbP = 2 int_0^infty x mathbbP(|a| > x),dx le 2 int_0^infty x p^x,dx < infty$$
which is a convergent integral.

Now if $f_k = a^2 1_ > k$, we have $|f_k| le a^2$ and $f_k to 0$ a.e. We have just shown $a^2$ is integrable, so by dominated convergence, $int f_k ,dmathbbP to 0$ as desired.