You have $x_{k-1} \leq t_k < 0 \leq t_{k+1} \leq x_{k+1}$, $x_k \in (x_{k-1},x_{k+1})$, and $x_{k+1} - x_{k - 1} < 2\delta$, so $|x_k| < 2\delta$.
I would probably have defined $k$ in terms of where the $x_i$'s change sign rather than where the $t_i$'s do, but your method works too.
In 3. we are given that $f$ and $g$ are Riemann integrable on $[a,b]$ and $g(x) \geqslant 0$ for all $x \in [a,b]$. We also are given that $f$ has a primitive $F$ implying that $f(x) = F'(x)$ for all $x \in [a,b]$.
Proof of the mean value theorem for integrals is, of course, straightforward under the stronger condition that $f$ is continuous. It is also true with this weaker condition that $f$ is a derivative. Let's first prove this carefully and then consider your final question where in one case $\mu = \inf f(x)$.
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = \inf_{x \in [a,b]}\, f(x)$ and $M = \sup_{x \in [a,b]}\, f(x)$. Since $mg(x) \leqslant f(x)g(x) \leqslant Mg(x)$ for all $x \in [a,b]$ we have
$$m\int_a^b g(x) \, dx \leqslant \int_a^b f(x) \, g(x) \, dx \leqslant M \int_a^b g(x) \, dx.$$
In the case where $\int_a^b g(x) \, dx = 0$ it is easy to show that we can choose any $\xi \in (a,b)$ and the theorem holds.
Otherwise take $\mu = \int_a^b f(x) \, g(x) \, dx / \int_a^b g(x) \, dx$. We know that $m \leqslant \mu \leqslant M$. If $m < \mu < M$, by the properties of infimum and supremum there exist $\alpha , \beta \in [a,b]$ such that
$$m < f(\alpha) < \mu < f(\beta) < M.$$
By Darboux's theorem $f$ (as a derivative) has the intermediate value property. Hence, there exists $\xi \in (\alpha,\beta) \subset [a,b]$ such that $f(\xi) = \mu$ and we are done.
Suppose however that $\mu = m$. Since $f(x) \geqslant m$ and $g(x) \geqslant 0$, we have
$$\int_a^b |f(x) - m| \, g(x) \, dx = \int_a^b (f(x) - m) \, g(x) \, dx = (\mu -m) \int_a^b g(x) \, dx = 0,$$
and it follows that $(f(x) - m) \, g(x) = 0$ almost everywhere. For this case where $\int_a^b g(x) \, dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $\xi \in (a,b)$ such that $f(\xi) = m$.
The case where $\mu = M$ is handled in a similar way.
Best Answer
We want to prove the following:
Although this is somewhat reminiscent of a mean value theorem for integrals, it's much simpler. Call $$ \int_a^b f(x) dx = I,$$ which exists since $f$ is integrable. It is very easy to show that $m(b-a) \leq I \leq M(b-a)$, and I take this for granted.
Then you can consider a function $g(x) = I - x(b-a)$ on the interval $[m,M]$. Now $g$ is a continuous function. We know $g(m) \geq 0$ and $g(M) \leq 0$, so by the intermediate value theorem there is some $\mu \in [m,M]$ such that $I - \mu(b-a) = 0$, or rather $I = \mu(b-a)$. $\diamondsuit$
The strength of the usual integral mean value theorem is that there is some $c$ so that the integral is given by $f(c)(b-a)$, in particular that it's given by a particular value of $f$.
We cannot hope to prove quite the same thing here. For instance, consider the function given by $0$ from $0$ to $1$ and $2$ from $1$ to $2$. Then the average value is $1$, which is not a value taken by the function.