# Riemann-Stieltjes integrable (Proof)

I would like to prove a version of the following theorem. Theorem : Let $I$ be a bounded interval , and let $f$ be a function which is uniformly continuous on $I$. Then $f$ is Riemann integrable (RI).

The theorem I want to prove: Theorem: Let $I$ be a bounded interval , and let $f$ be a function which is uniformly continuous on $I$. Then $f$ is Riemann-Stieltjes integrable (RSI).

Proof: Since f is uniformly continuous on a bounded interval, it is bounded. Say $M \in R^+$ such that $\forall x \in I$, we have that $\vert f(x)\vert \leqslant M$. By the definition of Uniform continuity,

$\forall \epsilon >0, \exists \delta>0$ such that

if $\vert x -y \vert < \delta$, then $\vert f(x)-f(y) \vert < \epsilon$ Now let $N \in Z^+$ such that $0< (b-a)/N <\delta$.. Let P be a a uniform partition with length of subintervals $(b-a)/N$.

Then by definition of upper RS sum: $$\overline \int_I f d\alpha \leq \sum_{k=1}^Nsup(f(x))\alpha[J_k]$$ & $$\underline \int_I f d\alpha \leq \sum_{k=1}^Ninf(f(x))\alpha[J_k]$$, with $\alpha: I \rightarrow$ R, and $\alpha[J_k]=\alpha(b_k)-\alpha(a_k)$.

Now, we have

$$0\leq \overline \int_I f d\alpha-\underline \int_I f d\alpha \leq \sum_{k=1}^Nsup(f(x))\alpha[J_k]-\sum_{k=1}^Ninf(f(x))\alpha[J_k]$$.. By this inequality we can use the properties of the sums

$$\sum_{k=1}^Nsup(f(x))\alpha[J_k]-\sum_{k=1}^Ninf(f(x))\alpha[J_k]=\sum_{k=1}^N\alpha[J_k](sup(f(x))-inf(f(x))$$

and therefore,

$$0\leq \overline \int_I f d\alpha-\underline \int_I f d\alpha \leq \sum_{k=1}^N\alpha[J_k](sup(f(x))-inf(f(x))$$.. And here where I am getting stuck, so I do not know if I made some mistake above or not, but anyway I am having an idea that since we have a uniform partition, then $\alpha[J_k]$ is constant , in particular,$\alpha[J_k]= \alpha((b-a)/N)$ for all $k$. Moreover we have chosen $N$ to be arbitrary so it is true for any $\alpha((b-a)/N)>0$ which means

$$0\leq \overline \int_I f d\alpha-\underline \int_I f d\alpha \leq \alpha((b-a)/N)\sum_{k=1}^N(sup(f(x))-inf(f(x))$$ AND HENCE, that will make

$$\overline \int_I f d\alpha-\underline \int_I f d\alpha=0$$

I appreciate any help with that..

Thank you.