Show that $\frac{d}{dt}\int_Sf(x,t)=\int_S\frac{\partial f}{\partial t}(x,t)$ for each $t\in I$ when $S$ is a fixed integrable set.

Question

Let be $S\subseteq\Bbb R^n$ a fixed integrable set and let be $f(x,t)$ scalar function of class $C^1$ defined in $S\times I$ where $I$ is an interval. So I ask to prove that the identity
$$
\frac{d}{dt}\int_Sf(x,t)=\int_S\frac{\partial f}{\partial t}(x,t)
$$
holds fore each $t\in I$. So to prove it I advanced the following argumentations. First of all thorugh the linearly if integral (is this correct?) I observed that
$$
\frac{d}{dt}\int_Sf(x,t):=\lim_{t\rightarrow t_0}\frac{\int_S f(x,t)-\int_Sf(x,t_0)}{t-t_0}=\lim_{t\rightarrow t_0}\frac{\int_S\big(f(x,t)-f(x,t_0)\big)}{t-t_0}=\lim_{t\rightarrow t_0}\int_S\Biggl(\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)
$$
for each $t_0\in I$ and so the statement follows directely showing that
$$
\lim_{t\rightarrow t_0}\int_S\Biggl(\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)=\int_S\Biggl(\lim_{t\rightarrow t_0}\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)=
$$
but unfortunately I did not able to do this. However by the mean value theorem (is this true?) I know that there must exist $\theta_t\in(0,1)$ such that
$$
\frac{\partial f}{\partial t}(x,t_0+\theta_t(t-t_0))=\frac{f(x,t)-f(x,t_0)}{t-t_0}
$$
so that
$$
\lim_{t\rightarrow t_0}\int_S\Biggl(\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)=\lim_{t\rightarrow t_0}\int_S\frac{\partial f}{\partial t}(x,t_0+\theta_t(t-t_0))
$$
and so I think that the resul follows showing that if $t\rightarrow t_0$ then $\theta_t\rightarrow 0$ but unfortunately I did not able to do this. Finally by the mean value integral theorem we know that for each $t\in I$ exists $\xi_t\in S$ such that
$$
\int_S\Biggl(\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)=\frac{f(\xi,t)-f(\xi,t_0)}{t-t_0}\cdot\text{vol}(S)
$$
so that
$$
\lim_{t\rightarrow t_0}\int_S\Biggl(\frac{f(x,t)-f(x,t_0)}{t-t_0}\Biggl)=\lim_{t\rightarrow t_0}\Biggl(\frac{f(\xi,t)-f(\xi,t_0)}{t-t_0}\cdot\text{vol}(S)\Biggl)=\\
\Biggl(\lim_{t\rightarrow t_0}\frac{f(\xi,t)-f(\xi,t_0)}{t-t_0}\Biggl)\cdot\text{vol}(S)=\frac{\partial f}{\partial t}(\xi,t_0)\cdot\text{vol}(S)
$$
so that the statement follows showing that
$$
\frac{\partial f}{\partial t}(\xi,t_0)\cdot\text{vol}(S)=\int_S\frac{\partial f}{\partial t}(x,t_0)
$$
but again I did not able to do this. So could someone help me, please? I point out I did NOT study Lebesgue integration theory so I courteously ask to do not use it, thanks.

Answer 1

By the mean value theorem, we have

$$\frac{f(x,t) - f(x,t_0)}{t-t_0} = \frac{\partial f}{\partial t}(x,\xi)$$ where $\xi$ is between $t$ and $t_0$ and may depend on $x$.

The easiest proof arises with the strongest hypothesis where $f$ is uniformly continuous on $S \times I$. In that case we have

$$\left|\int_S \frac{f(x,t)- f(x,t_0)}{t-t_0} -\int_S \frac{\partial f}{\partial t}(x,t_0)\right| \leqslant \int_S \left|\frac{\partial f}{\partial t}(x,\xi)- \frac{\partial f}{\partial t}(x,t_0) \right|, $$

By uniform continuity, for any $\epsilon > 0$ there exists $\delta > 0$ depending only on $\epsilon$, such that if $|x_1 - x_2| < \delta $ and $ |t_1 - t_2| < \delta$, then

$$\left|\frac{\partial f}{\partial t}(x_1,t_1)- \frac{\partial f}{\partial t}(x_2,t_2) \right| < \frac{\epsilon}{vol(S)}$$

Hence, if $|t-t_0| < \delta$ it follows that $|\xi - t_0| < \delta$ and

$$\left|\int_S \frac{f(x,t)- f(x,t_0)}{t-t_0} -\int_S \frac{\partial f}{\partial t}(x,t_0)\right| < \epsilon$$

Whence,

$$\left.\frac{d}{dt} \int_s f(x,t)\right|_{t = t_0} = \int_S \frac{\partial f}{\partial t} (x, t_0)$$