This was a problem in my analysis textbook-
Suppose that $f(x)\le g(x)\le h(x)$ for all $x\in[a,b]$ and that $\int_{a}^{b}f$ and $\int_{a}^{b}h$ exists and are equal. Prove that $\int_{a}^{b}g$ exists and equals $\int_{a}^{b}f =\int_{a}^{b} h$.
Attempted proof:
We first prove that $g$ is integrable on $[a,b]$. We claim that for any partition $P$ of $[a,b]$, we have $L_P(f)\le L_P(g)$ and $U_P(g) \le U_P(h)$. Our claim is easy to verify: If $J$ be any subinterval of $[a,b]$, then $f(x)\le g(x)$ on $J$. Clearly, $m_J(f) =\inf \{f(x) \mid x \in J\} \le g(x)$ for all $x \in J$. It follows that $m_J(f)$ is a lower bound for the set $\{g(x) \mid x\in J\}$. Hence, $m_J(f) \le m_J(g)$. It immediately follows by the definition of lower sum that $L_P(f)\le L_P(g)$ for any partition P. It can be verified similarly that $U_P(g) \le U_P(h)$.
Let $\varepsilon >0$ be given. Since $f$ is integrable on $[a,b]$, there exists a partition $P_1$ of $[a,b]$ such that $U_{P_1}(f)-L_{P_1}(f)<\frac{\varepsilon}{2}$. Likewise, since $h$ is integrable on $[a,b]$, there exists a partition $P_2$ of $[a,b]$ such that $U_{P_1}(h)-L_{P_1}(h)<\frac{\varepsilon}{2}$. Let $P=P_1 \cup P_2$. Then we have the following: $$U_P(f)-L_P(f)\le \frac{\varepsilon}{2}$$ and $$U_P(h)-L_P(h)\le \frac{\varepsilon}{2}$$
Now, we consider the following:
$$\begin{align} U_P(g)-L_P(g) &\le U_P(h)-L_P(h) \\ & < \left( L_P(h) + \frac{\varepsilon}{2} \right) - \left( U_P(f) -\frac{\varepsilon}{2} \right) \\ & = \left( L_P(h) - U_P(f) \right) + \varepsilon \end{align}$$
To complete the proof, we notice that $L_P(h)\le \int_{a}^{b} h$ and $\int_{a}^{b} f \le U_P (f)$. Hence, it follows that $L_P(f) - U_P(f) \le \int_{a}^{b} h -\int_{a}^{b} f =0$. It follows from this that $$U_P(g)-L_P(g) < \varepsilon$$ Hence, $g$ is integrable on $[a,b]$.
Is this proof correct? Or did I go wrong somewhere? Alternative proofs are appreciated.
Here is an alternative proof that seems easier to me.
By your own argument, we have $L(P,f) \leqslant L(P,g) \leqslant U(P,g) \leqslant U(P,h)$ for any partition $P$.
Thus, since $f$ and $h$ are Riemann integrable,
$$\int_a^bf = \sup_{P}L(P,f) \leqslant \sup_{P} L(P,g)= \underline{\int}_a^bg \\\leqslant \overline{\int_a^b}g = \inf_{P}\,\,U(P,f) \leqslant \inf_{P} \,\,U(P,h)= \int_a^bh$$
Since the integrals of $f$ and $h$ are equal, it follows that the upper and lower Darboux integrals of $g$ must be equal,
$$\underline{\int}_a^bg = \overline{\int_a^b}g, $$
whence, $g$ is Riemann integrable with
$$\int_a^b f = \int_a^b g = \int_a^b h$$