Differentiation under the integral sign in higher dimensions

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Let's say I have a function $f : \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. Are there conditions under which the following holds

$$ \frac{\partial}{\partial\mathbf{x}^T}\int_\Omega f(\mathbf{x},t) \,{\rm d} t = \int_\Omega \frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t) \,{\rm d} t,$$

where $\Omega\subset\mathbb{R}$ is some open set? In this case, both the LHS and RHS quantities are in $\mathbb{R}^{n \times n}$.

For $n=1$ there are results that follow from Lebesgue Dominated Convergence, but I am unaware of results in higher dimensions. Intuitively, it seems like the extension shouldn't be too complicated.

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The goal will be to reduce to 1D, so that we can apply that standard differentiation under the integral sign, which I will call DIS. See wikipedia for quick reference on this result.

Assume the following:

  1. For almost all $t\in\Omega$, $\frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)$ exists for all $\mathbf{x}\in\mathbb{R}^n$

  2. $f(\mathbf{x},t)$ is integrable w.r.t $t$ for all $\mathbf{x}\in\mathbb{R}^n$

  3. $||\frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)||_2\leq g(t)$ for all $\mathbf{x}\in\mathbb{R}^n$ and almost all $t\in\Omega$, where $g$ is an integrable function

From 3, we can conclude the following for $i,j=1,\ldots,n$: $$\Big(\frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)\Big)_{ij}\leq||\frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)||_{max}\leq||\frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)||_2\leq g(t)$$ for all $\mathbf{x}\in\mathbb{R}^n$ and almost all $t\in\Omega$. Now, apply DIS to yield the following: $$\Big(\frac{\partial}{\partial\mathbf{x}^T}\int_\Omega f(\mathbf{x},t)\ dt\Big)_{ij}=\Big(\int_\Omega \frac{\partial}{\partial\mathbf{x}^T}f(\mathbf{x},t)\ dt\Big)_{ij}$$

Then we can conclude that the two matrices in question are equal via equality of their elements.