1

Fix an arbitrary \(t_0 \in [a,b]\) and we show that \(F\) is continuous at \(t_0\). Let \((t_n)\) be a sequence of numbers in \([a,b]\) that converges to \(t_0\).

Since \(f\) is continuous in its second coordinate, for each \(x\in X\), we have \[\lim_{n\to\infty} f(x,t_n) = f(x,t_0).\]

As for each \(t_n\), we have \(\left\lvert f(x,t_n)\right\rvert\leq g(x)\) we can apply dominated convergence to show that \[ F(t_0) = \lim_{n\to\infty} F(t_n) = \lim_{n\to\infty} \int f(x,t_n)\ d\mu(x). \]

2

Fix \(t\in[a,b]\).

For any sequence \((t_n)\subseteq [a,b]\) with \(t_n\neq t\), we have \[ \frac{\partial f}{\partial t}(x,t) = \lim_{t_n\to t} \frac{f(x,t_n)-f(x,t)}{t_n-t} \] so that \(x\mapsto \frac{\partial f}{\partial t}(x,t)\) is a pointwise limit of measurable functions and therefore measurable.

For any \(x\in X\) we can apply mean value theorem to \(t\mapsto \frac{\partial f}{\partial t}(x,t)\) to obtain existence of \(s\) between \(t_0\) and \(t\) such that \[ f(x,t) - f(x,t_0) = (t - t_0)\frac{\partial f}{\partial t}(x,s) \] then we have \[ \left\lvert f(x,t)\right\rvert \leq \left\lvert f(x,t_0)\right\rvert + \left\lvert t-t_0\right\rvert g(x) \] which means that \(x\mapsto f(x,t)\) is integrable for each \(t\), not just \(t_0\).

Now for each \(n\), \[ \frac{F(t_n) - F(t)}{t_n - t} = \int \frac{f(x,t_n) - f(x,t)}{t_n-t}\ d\mu(x) \] and the integrated is dominated by \(g\), so by dominated convergence, \[ \frac{dF}{dt}(t) = \lim_{n\to\infty} \frac{F(t_n) - F(t)}{t_n - t} = \int \frac{\partial f}{\partial t}(x,t)\ d\mu(x). \]