If $P(|a|>k)\leq p^k$ then $\lim_{k\to \infty}\int_{\{|a|>k\}}a^2d\mathbb{P} = 0$? 
Let $p \in (0,1)$. If $P(|a|>k)\leq p^k$ then $$\lim_{k\to \infty}\int_{\{|a|>k\}}a^2d\mathbb{P} = 0?$$

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.
 A: Yes, this is true. Break the sum up as:
\begin{align*}
\int_{|a|>k}a^2 \ d \mathbb{P}&=\sum_{n=k}^\infty  \int_{n<|a|\leq n+1} a^2 \ d \mathbb{P}\\
&\leq \sum_{n=k}^\infty (n+1)^2 \mathbb{P}(n<|a|\leq n+1)\\
&\leq \sum_{n=k}^\infty (n+1)^2 p^n.
\end{align*}
Now the sum above is convergent for $p \in (0,1)$, so in particular the tail sums have to be going to zero.
A: A variation on kccu's solution uses the trick that for any random variable $a$ and any $q > 0$, you can write
$$\int |a|^q\,d\mathbb{P} = q \int_0^\infty x^{q-1} \mathbb{P}(|a| > x)\,dx.$$
To prove this, write $\mathbb P(|a| > x) = \int 1_{|a| > x}\,d\mathbb{P}$ and interchange the integrals (which is justified since everything is nonnegative).  Then note that $\int_0^\infty q x^{q-1} 1_{|a| > x}\,dx = \int_0^{|a|} q x^{q-1}\,dx = |a|^q$ by the fundamental theorem of calculus.
With $q=2$ this reads
$$\int a^2\,d\mathbb{P} = 2 \int_0^\infty x \mathbb{P}(|a| > x)\,dx \le 2 \int_0^\infty  x p^x\,dx < \infty$$
which is a convergent integral.
Now if $f_k = a^2 1_{|a| > k}$, 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, $\int f_k \,d\mathbb{P} \to 0$ as desired.
