Let $\psi$ be defined by$$\psi(s):=\int_{[a,b]}K(s,t)\varphi(t)d\mu_t$$ where $\varphi\in L_2[a,b]$ and $K\in L_2([a,b]^2)$. Kolmogorov-Fomin's proves the belonging of $\psi$ to $L_2[a,b]$ by showing that $\forall s\in[a,b]\quad|\psi(s)|^2\le\|\varphi\|^2\int_{[a,b]}|K(s,t)|^2d\mu_t$, and the fact that $\int_{[a,b]}\int_{[a,b]}|K(s,t)|^2d\mu_td\mu_s$ exists because $K\in L_2([a,b]^2)$.
I think that such majorations to prove summability work if $\psi$ is measurable, but how can we see that it is? Thank you very mcuh!
One possibility is the following:
Decompose $K =K_+ - K_-$ and $\phi =\phi_+ -\phi_-$, where $K_\pm$ are the positive and negative parts of $K$.
Then your integrand is
$$ (K_+ - K_-)\cdot (\phi_+-\phi_-)= K_+ \phi_+ + K_- \phi_- -(K_+ \phi_- + K_- \phi_+). $$
Now each of the contents of the two brackets is a measurable, nonnegative function.
Fubini's (or Fubini-Tonelli, if you want) Theorem implies that
$$ [a,b]\to [0,\infty], x \mapsto \int f(x,y) \, d\mu(y) $$ is a measurable function (at least if $\mu$ is sigma finite), if $f$ is a nonnegative, measurable function (wrt. the product sigma algebra).
Use this on the two brackets. Then the proof shows that the inner integral of each of the two brackets alone is finite almost everywhere (this relates to your other question, which I just answered), so that you can subtract both of these integrals and obtain a measurable function.
If your functions are complex valued, first decompose into real and imaginary part.