I am so confused on how to do this, my professor made it sound simple and I'm sure it is, but I cannot figure this out:
Let $g\in \mathcal{C}^1(\mathbb{R}^{N+1})$ and define $f(x)=\int_a^b g(x,y)dy$ for $x\in\mathbb{R}^N$. Use the Dominated Convergence Theorem to show that for every $j=1,..., N$ we have
$$\frac{\partial}{\partial x_j}f(x) =\int_a^b\frac{\partial}{\partial x_j} g(x,y)dy.$$
So, I know that
$$\frac{\partial}{\partial x_j}f(x) = \frac{\partial}{\partial x_j}\int_a^b g(x,y)dy$$ and then the DCT will let us swap the integral with the partial derivative, but how on earth do you use the DCT here? [If $f_n\in L^1(\mathbb{R}^N)$ converges almost everywhere to a function $f$ and there exists $h\in L^1(\mathbb{R}^N)$ such that $|f_n|\leq h$ for all $n$, then $f\in L^1(\mathbb{R}^N)$ and $f_n\to f$ in norm.] Any help would be much appreciated, thank you! Also, we did not really use measure much in class, as weird as that sounds.
We’re going to use the mean value theorem construct a dominating function over a sequence of functions which converges to $\partial_xf$.
First, satisfy yourself that $f_n(x,y)=n(f(x+\frac{1}{n},y)-f(x,y))$ converges to $\partial_x f$
Applying the mean value theorem, notice that for some $a\in(x,x+\frac{1}{n})$, $\partial_xf(a,y)=f_n(x,y)$. This implies that $|f_n(x)|\leq|\sup_x \partial_x f(x,y)|$
We can therefore choose the constant function $g(x,y)= |\sup_x \partial_x f(x,y)|$. This is integrable because constant functions are integrable over finite intervals
I’ll leave the details to you, but apply the dominated convergence theorem to show that $\lim_{n\to\infty} \int f_n(x,y)= \int \lim_{n\to\infty} f_n(x,y)$