Problem: Let $f,g,h:[a,b]\to\mathbb{R}$ satisfy $f(x)\leq g(x)\leq h(x), \forall x \in[a,b]$. Suppose that $f$ and $h$ are integrable functions on $[a,b]$. Suppose further that $\int_a^bf=\int_a^bh$. Prove that $g$ is integrable on $[a,b]$.
My solution/proof: Let $f:[a,b]\to\mathbb{R}$ be a bounded function, and P be a partition of $[a,b]$. $P=(x_0,x_1,...,x_n)$. $$M_r=sup_{x\in[x_{r-1},x_r]}f(x)$$$$m_r=inf_{x\in[x_{r-1},x_r]}f(x)$$.
Upper Darboux sum: $$U(P,f)=\sum_{r=1}^nM_r(x_r-x_{r-1})$$
Lower Darboux sum: $$L(P,f)=\sum_{r=1}^{n}m_r(x_r-x_{r-1})$$
Now $$sup{U(P,f): P \in P[a,b]}=\int_{a}^{\bar{b}}f$$ $$inf{L(P,f): P \in P[a,b]}=\int_{\underline{a}}^{b}f$$
If $P[a,b]$ is any partition of $[a,b]$, then $f$ is integrable if $$\int_{a}^{\bar{b}}f=\int_{\underline{a}}^{b}f$$
Here, $f,g,h:[a,b]\to\mathbb{R}$ and $f\leq g\leq h, \forall x\in[a,b]$. $f,h$ are integrable, so $f,g$ are bounded and hence, $g$ is bounded. So, $$\int_{a}^{\bar{b}}f=\int_{\underline{a}}^{b}f\:\:\:\:\: and \:\:\:\: \int_{a}^{\bar{b}}h=\int_{\underline{a}}^{b}h.$$
Since $f\leq g \leq h, \forall x\in[a,b]$, then for any partition $P\in P[a,b]$ $$L(P,f)\leq L(P,g)$$ $$U(P,g)\leq U(P,h)$$ $$inf_{P \in P[a,b]}L(P,f)\leq inf_{P \in P[a,b]}L(P,g)$$ $$supU_{P \in P[a,b]}(P,g)\leq supU_{P \in P[a,b]}(P,h)$$
So $$\int_{\underline{a}}^{b}f \leq \int_{\underline{a}}^{b}g$$ $$\int_{a}^{\bar{b}}g \leq \int_{a}^{\bar{b}}h$$ But $$\int_{\underline{a}}^{b}f = \int_{a}^{\bar{b}}h$$ So $$\int_{a}^{\bar{b}}g = \int_{\underline{a}}^{b}g$$
Therefore, $g$ is integrable on $[a,b]$.
Is the proof correct?