Differentiating under the Integral with Carathéodory.

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………..

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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 ?