Browsing through Tao's notes on harmonic analysis and trying to understand his proof of the Riesz-Thorin theorem he states nonchalantly that for $F_0,F_1,G_0,G_1,a,b$ simple (functions map from $X$ being a measure space to $\mathbb{C}$) and $T$ a linear operator from $X$ to $Y$ where $X$ and $Y$ are measure spaces the function defined as
$$H(z) = \int_X T(F_0^{(1-z)}F_1^z a)(G_0^{(1-z)}G_1^z b) d \mu$$
is entire since all functions inside the integral are simple and $T$ is linear. Now is this as clear as he makes it seem? In the real case we know by the fundamental theorem we know that the integral over an integrable function is absolutely continuous and hence differentiable almost everywhere. But is this somewhat analogously true in the complex case? Because the only thing we know about the integrand here is that it is integrable as the product of simple functions subjected to a linear operator.
Every simple function $f$ is of the form $f = \sum_{k=1}^n f_k \chi_{A_k}$, where $\chi_{A_k}$ are the indicator functions of disjoint measurable sets $A_k$, and $f_K$ are just some complex numbers. Then $f^z = \sum_{k=1}^n f_k^z \chi_{A_k}$, where $f_k^z = \exp(z\log(f_k))$ is an entire function (you have to make a choice of $\log(f_k)$, but which doesn't matter if $z \in \{0,1\}$). I'll be lazy and just consider the expression $\int_X T(f^z)g^z d\mu$. By linearity of $T$ and the integral, we can pull out the finite sum to obtain
$$ \int_X T(f^z)g^z d\mu= \int_X T\left(\sum_{k=1}^n f_k^z \chi_{A_k}\right)\left(\sum_{l=1}^m g_l^z \chi_{B_l}\right) d\mu = \sum_{k=1}^n \sum_{l=1}^m f_k^z g_l^z \int_X T(\chi_{A_k})\chi_{B_l} d\mu.$$
Now it's clear that this expression is entire, since $\int_X \chi_{A_k}\chi_{B_l}$ is constant, so this is really just a finite sum of exponentials.
Terry Tao's expression is a bit more involved, but in principle everything works in just the same way, because products of simple functions are still simple functions.