Measurability of inner integral $x \mapsto \int f(x,y)\, d\mu(y)$

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!

• I allowed myself to change the title of the question. When I read the title, I thought you were interested in measurability of $F(H)$ as a subset of some Hilbert space, where $F$ is a Fredholm operator defined on some Hilbert space $H$. – PhoemueX Dec 7 '14 at 15:12
• Thank you for the precisation! – Self-teaching worker Dec 7 '14 at 17:35

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.

• I'm not sure to understand one thing: how does Fubini's theorem imply the measurability of $x\mapsto\int f(x,y)d\mu_y$ and is it valid for any measurable domain instead of $[a,b]$? Thank you again!!! – Self-teaching worker Dec 7 '14 at 17:43
• Yes, Fubini's Theorem (or the Fubini Tonelli theorem) states that if $\mu,\nu$ are sigma-finite measures and if $f : X\times Y \to [0,\infty]$ is measurable w.r.t. the produce $\sigma$-algebra, then $\int f(x,y) \, d\nu(y) \in [0,\infty]$ exists for all $y$ and the map $x \mapsto \int f(x,y)\, d\nu(y)$ is measurable. The same of course holds for $y \mapsto \int f(x,y) \, d\mu(x)$. Hence, this is also true for other domains than $[a,b]$. For this version of Fubini's theorem, see e.g. Folland's "Real Analysis", or Rudin's "Real & Complex Analysis". See also planetmath.org/tonellistheorem – PhoemueX Dec 7 '14 at 18:30