Let $f(x, y)$ be a real valued function defined on the real domain $X \times Y$ which is integrable in the Riemann or Lebesgue sense (or any other ?) on this domain.
Further, let the partial derivative $\frac {\partial{f(x, y)}} {\partial{y}}$ be defined throughout the domain.
We want to show that $\frac {\partial{\int f(x, y)} dx} {\partial{y}} =\int \frac {\partial{f(x, y)}} {\partial{y}} dx$ for all $y \in Y$.
I'd appreciate feedback on my proof attempt which follows...........
Take the Carathéodory definition of differentiability (see e.g. Concerning Carathéodory's criteria of differentiability and a proof that differentiable implies continuous ), clearly equivalent to the usual definition.
By this definition, $\frac {\partial{f(x, y)}} {\partial{y}}$ is defined throughout the domain means that for all $x, y, y_0$ in the domain there is a function $\psi$ such that $f(x, y) = f(x, y_0) + (y - y_0)\psi(x, y)$ where $\psi(x, y)$ is continuous at $y_0$ and then $\frac {\partial{f(x, y)}} {\partial{y}}|_{y_0} = \psi(x, y_0)$.
Since $f$ is integrable then $\int{f(x, y)}dx = \int{f(x, y_0)} + (y - y_0)\psi(x, y) dx $
And, $f(x, y_0)$ is x-integrable too, so $(y - y_0)\psi(x, y)$ is x-integrable and
$\int (y - y_0)\psi(x, y)dx = \int{f(x, y)}dx - \int{f(x, y_0)} dx= (y - y_0)\int \psi(x, y)dx$
So, $\int{f(x, y)}dx = \int{f(x, y_0)}dx + (y - y_0) \int \psi(x, y) dx $
Then provided $\int \psi(x, y) dx $ is continuous at $y = y_0$ for all $y_0$ this is the Carathéodory definition that $\int{f(x, y)}dx $ is y-differentiable at $y_0$ and that $\frac {\partial{\int f(x, y)} dx} {\partial{y}}|_{y_0} = \int \psi(x, y_0) dx =\int \frac {\partial{f(x, y)}} {\partial{y}}|_{y=0} dx$ for all $y_0 \in Y$.
But I don't see how to prove continuity ?