Alternative proof for Riemann integration Theorem I already have a proof of the following theorem and I was wondering if there is another kind of proof, (just being curious). 
Suppose $f$ is Riemann integrable and $h$ is non-decreasing and
non-negative. Let $F(x) = \int_x^b f$. If $A \leqslant F(x) \leqslant
B$ for all $x \in [a,b],$ then $h(b)A \leqslant \int_a^b f h \leqslant
 h(b)B$.
Proof:
Taking any partition $P = (x_0,x_1, \ldots, x_n)$ consider the Riemann sums
$$S_P = \sum_{k=1}^n f(x_k) h(x_k)(x_k -x_{k-1}), \\ S_{P,j} = \sum_{k = j}^n f(x_k)(x_k - x_{k-1}),$$
which converge to $\int_a^b fh $ and $\int_{x_{j-1}}^b f$.
Since $f(x_k)(x_k - x_{k-1}) = S_{P,k} - S_{P,k-1}$ we have 
$$S_P = \sum_{k=1}^n h(x_k)(S_{P,k} - S_{P,k-1}) \\ = h(x_1)S_{P,1} + (h(x_2) - h(x_1))S_{P,2} + \ldots (h(x_n) - h(x_{n-1}))S_{P,n} $$
Let $\hat{A}$ and $\hat{B}$ be the upper and lower bounds for the finite set $\{S_{P,k}\}$. Since $h$ is non-decreasing $h(x_k) - h(x_{k-1}) \geqslant 0$ and
$$\hat{A} h(b) = \hat{A} h(x_n) \leqslant S_P \leqslant \hat{B} h(x_n) = \hat{B} h(b).$$
As the number of partition points increases the sum $S_P$ converges to $\int_a^b fh $ and it can be shown that $\hat{A} \to A$ and $\hat{B} \to B.$
Therefore,
$$h(b)A \leqslant \int_a^b f h \leqslant h(b)B$$
Note: This proof was made by RRL user :)
 A: Here is another way to prove this lemma without using Riemann sums.  Again we have weak conditions where $f$ is Riemann integrable and $h$ is non-decreasing and non-negative -- but neither is assumed to be continuous everywhere.
Defining $F(x) = \int_x^b f(t) \, dt$, since $f$ is integrable we have that $F$ is absolutely continuous and $F'(x) = -f(x)$ almost everywhere.  We can apply integration by parts for Riemann-Stieltjes integrals to obtain
$$-\int_a^b f(x)h(x) \, dx = \int_a^b h \, dF = F(b)h(b) - F(a)h(a) - \int_a^b F \, dh.$$
Since $F(b) = 0$ it follows that
$$\tag{1}\int_a^b f(x)h(x) \, dx =  F(a)h(a) + \int_a^bF \,dh.$$
Since $F$ is continuous it is bounded on $[a,b]$.  Let $A = \inf_{x \in [a,b]}F(x)$ and $B = \sup_{x \in [a,b]}F(x)$.
Since $A \leqslant F(x) \leqslant B$ for all $x \in [a,b]$ and $h$ is non-decreasing we have
$$\tag{2}A(h(b) - h(a)) \leqslant \int_a^b F\, dh \leqslant B(h(b) - h(a)).$$
Hence, from (1) and (2) it follows that
$$Ah(b) - Ah(a) + F(a) h(a) \leqslant \int_a^b f(x) h(x) \, dx \leqslant Bh(b) - Bh(a) + F(a) h(a),$$
and
$$Ah(b) + (F(a) - A)h(a) \leqslant \int_a^b f(x) h(x) \, dx \leqslant Bh(b) -(B- F(a))h(a).$$
Since $A \leqslant F(a) \leqslant B$ and $h$ is non-negative, we have $(F(a)-A)h(a) \geqslant 0$ and $(B- F(a))h(a) \geqslant 0$.
Therefore,
$$Ah(b) \leqslant \int_a^b f(x) h(x) \, dx \leqslant Bh(b).$$
