I have taken an introductory course on measure theory where I learned about the Borel-Cantelli theorem but I wonder whether there is a lebesgue integrable version. Given an uncountable collection of independent events $E\_{t \in \mathbb{R}\_+}$, 


$$ \int\_0^{\infty} P(E\_t) dt <\infty \implies P( E\_t\quad i.o. )=0\tag{1}$$


$$ \int\_0^{\infty} P(E\_t) dt =\infty \implies P( E\_t\quad i.o. )=1\tag{2}$$ 


**Note**: i.o. here means infinitely often. I'd like to add that this question is motivated by a problem I encountered in statistical physics. 



Integrable version of the Borel-Cantelli theorem?
The second Borel-Cantelli lemma has the additional condition that the events are *mutually independent*. This requirement becomes problematic for an uncountable index set. For example, suppose that $\{E\_t\}\_{0\le t\le 1}$ is a collection of mutally independent events such that $(\omega,t)\mapsto 1\_{E\_t}(\omega)$ is measurable. (This is needed for the integrations you envision.) Then 
$$
\int\_a^b 1\_{E\_t}\,dt =\int\_a^bp(t)\,dt,\qquad a.s.,
$$
where $p(t):=P(E\_t)$. (To see this compute the variance of $\int\_a^b 1\_{E\_t}\,dt$.) From this and the Lebesgue density and differentiation theorems it follows that $P(E\_t)$ is $0$ or $1$ for a.e. $t$. Thus mutual independence and measurability hold only in trivial situations.


