Riemann Integration and Squeeze Theorem Let $[a,b]\subseteq \mathbb{R}$ be a non-degenerate closed bounded interval, and let$f,g,h:[a,b]\to\mathbb{R}$ be functions. Suppose that $f$ and $h$ are integrable, and that $\int_a^bf(x)dx=\int_a^bh(x)dx$. Prove that if $f(x)\leq g(x) \leq h(x)$ for all $x\in[a,b]$, then $g$ is integrable and $\int_a^bg(x)dx=\int_a^bf(x)dx$.
I am not allowed to use squeeze theorem. Only use the definition of Riemann integral.
 A: Write $U(P, g)$ and $L(P, g)$ to denote the lower and upper sums for function $g$ over specified partition $P$ of $[a,b]$. Pick a partition $P_1$ and $P_2$ such that
$$ U(P_1, h)-\int_a^b h(x)dx<\epsilon$$
and 
$$ \int_a^b f(x)dx-L(P_2, f)<\epsilon$$
This is possible because both $f$ and $h$ are Riemann integrable. Let $P = P_1\cup P_2$ (common refinement of $P_1$ and $P_2$). Observe that $f(x)\le g(x)\le h(x)$ imply
$$ U(P, g)-L(P, g)\le U(P, h) - L(P, f)\le U(P_1, h)-L(P_2, f) < 2\epsilon $$
This shows that $g$ is Riemann integrable. Can you proceed from here?
A: Since $f,h\in \mathcal{R}[a,b]$, for a given $\epsilon>0$, there exists a $\delta_\epsilon>0$ such that for any tagged partition $\dot{\mathcal{P}}$ of $[a,b]$ with $||\dot{\mathcal{P}}||<\delta_\epsilon$, 
\begin{equation}
\left|S(f;\dot{\mathcal{P}})-\int_a^b f\right|<\epsilon,
\end{equation}
and
\begin{equation}
\left|S(h;\dot{\mathcal{P}})-\int_a^b h\right|<\epsilon.
\end{equation}
Since for the partition $\dot{\mathcal{P}}$
\begin{equation}
S(f;\dot{\mathcal{P}}) \leq S(g;\dot{\mathcal{P}}) \leq S(h;\dot{\mathcal{P}}),
\end{equation}
we have
\begin{equation}
\int_a^b f-\epsilon < S(f;\dot{\mathcal{P}}) \leq S(g;\dot{\mathcal{P}}) \leq S(h;\dot{\mathcal{P}})<\int_a^b h+\epsilon
\end{equation}
and by using the hypothesis $\displaystyle \int_a^b f=\int_a^b h$, one obtains
\begin{equation}
\left|S(g;\dot{\mathcal{P}})-\int_a^b f\right| <\epsilon.
\end{equation}
Therefore, $g\in\mathcal{R}[a,b]$ with
\begin{equation}
\int_a^b g=\int_a^b f.
\end{equation}
This completes the proof.
