Changing order of integration/expecation, probability problem Consider the ON-sequence $\{\varphi_{k}\}_{k}\in L^{2}(\mathbb{R})$ 
and let
$$I_{k}\in \mathrm{Bernoulli}(\lambda_{k}),\;\sum_{k=1}^\infty \lambda_{k}<\infty$$
$ \{I_k\} $ are all independent. Also let $A$ be a bounded intervall in $\mathbb{R}$
Is it allowed to change order of integration in the following manner?
$$\mathbb{E}\left[\intop_A \intop_{A^c}\sum_{k=1}^\infty \sum_{j=1}^\infty I_k I_j \varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y) \, dx \, dy\right]=$$
$$\intop_{A}\intop_{A^{c}}\sum\limits _{k=1}^{\infty}\sum\limits _{j=1}^{\infty}\mathbb{E}\left[I_{k}I_{j}\right]\varphi_{k}(x)\overline{\varphi_{j}(x)}\overline{\varphi_{k}(y)}\varphi_{j}(y)dxdy$$
This is how far i have come on my own:
I would like to invoke Borell Cantelli lemma, which says that the sum in the integral is finite a.s.. However for each $\omega\in\Omega$ outside a zero-set the number of terms will depend on $\omega$ so there is no single function that bounds the integrand for almost all $\omega$ and there for we may not move the expectation inside the integral.
 A: Consider $$S=\mathbb{E}\left[\intop_{A}\intop_{A^{c}}\sum\limits _{k=1}^{\infty}\sum\limits _{j=1}^{\infty}\left|I_{k}I_{j}\varphi_{k}(x)\overline{\varphi_{j}(x)}\overline{\varphi_{k}(y)}\varphi_{j}(y)\right|dxdy\right].$$
The sum $S$ of this series of nonnegative terms always exists in $\mathbb R_+\cup\{+\infty\}$. If $S$ is finite, Fubini theorem applied to the product measure of $\mathbb P$, two Lebesgue measures and two counting measures, ensures that every possible integral exists. 
To show that $S$ is indeed finite, note that $\mathbb{E}(I_{k}I_{j})=\lambda_k\lambda_j$ for every $k\ne j$ and $\mathbb{E}(I_{k}^2)=\lambda_k$ for every $k$. Hence, for every $A$,
$$S\leqslant\sum_{k=1}^{\infty}\lambda_k\left(\int_{\mathbb R}\left|\varphi_{k}(x)\right|^2dx\right)^2+\sum_{k\ne j}\lambda_k\lambda_j\left(\int_{\mathbb R}\left|\varphi_{k}(x)\varphi_{j}(x)\right|dx\right)^2,
$$
and, for every $(k,j)$,
$$
\left(\int_{\mathbb R}\left|\varphi_{k}(x)\varphi_{j}(x)\right|dx\right)^2\leqslant\int_{\mathbb R}\left|\varphi_{k}(x)\right|^2dx\cdot\int_{\mathbb R}\left|\varphi_{j}(x)\right|^2dx=1,
$$
hence
$$
S\leqslant\sum_{k=1}^{\infty}\lambda_k+\sum_{k\ne j}\lambda_k\lambda_j=\sum_{k=1}^{\infty}\lambda_k(1-\lambda_k)+\left(\sum_{k=1}^{\infty}\lambda_k\right)^2\leqslant\sum_{k=1}^{\infty}\lambda_k++\left(\sum_{k=1}^{\infty}\lambda_k\right)^2,
$$
which is finite by hypothesis, QED.
A: (Edit: this answer is no longer relevant to the question, following OP's edits.)
Nevermind the expectation. You have
$$\int\int\sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j\varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y)\,dx\,dy = \sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j \delta_{kj}^2 = \sum_{j=1}^\infty I_j.$$
Now taking expectation gives
$$\mathbf{E}\int\int\sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j\varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y)\,dx\,dy = \sum_{j=1}^\infty \lambda_j.$$
