Sufficient conditions for $z \to \int_\mathbb{R} h(z,x) \,d\mu(x)$ to be analytic

Question

Let $\mu$ be a measure on $\mathbb{R}$ and $G$ an open subset of $\mathbb{C}$. Every function $h \,:\, G\times\mathbb{R} \to \mathbb{C}$ then gives rise to a function $$ F_h \,:\, G \to \mathbb{C} \,:\, z \to \int_\mathbb{R} h(z,x) \,d\mu(x) $$ provided that $x\to h(z,x)$ is $\mu$-integrable for every $z \in G$.

I'm looking for sufficient conditions for $h$ which guarantee that $F_h$ is analytic on $G$.

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One sufficient condition seems to be that

1. $z \to h(z,x)$ is analytic on $G$ for all $x \in \mathbb{R}$ and
2. $h$ is $\lambda^2\times\mu$-measurable ($\lambda^2$ being the lebesgue measure on $\mathbb{C}$).

In that case, if $\gamma \,:\, \mathbb{R} \to \mathbb{C}$ is $\lambda$-measurable, then $(t,x) \to h(\gamma(t),x)$ is $\lambda\times\mu$ measurable and thus Fubini's theorem yields that $\int_{\partial B} \int_\mathbb{R} h(z,x) \,dx \,dz = \int_{\mathbb{R}} \int_{\partial B} h(z,x) \,dz\,dx = 0$ for every open ball $B \subset G$.

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I'd like to know if I'm missing anything in my proof above (the rather weak requirements make me a bit suspicious), and wheather there are other sufficient conditions for $F_h$ to be analytic.

Answer 1

Your proof looks fine to me, but I would argue from Weyl's lemma (holomorphic version):

If $F$ is locally integrable in $G$ and $\int_G F\varphi_{\bar z}\,d\lambda^2=0$ for all $\varphi\in C^\infty_c(G)$, then $F$ is holomorphic in $G$.

(This is if and only if, of course.)

So, all that needs to be justified is that $$\int_G \left(\int_{\mathbb R}h(z,x)\,\varphi_{\bar z}\,d\mu(x)\right)\,d\lambda^2 = \int_{\mathbb R} \left(\int_G h(z,x)\,\varphi_{\bar z}\,d\lambda^2 \right)\,d\mu(x)$$ for which you use Fubini.