# Riemann integral property proof Completely lost. I know I have to use the sup and it is an epsilon delta proof, but other than that, I am confused.

For any Riemann function (i.e., for any partition of $\;[a,b]\;$):
$$\sum_{||\Delta_x||\to 0}(cf)(c_i)(x_i-x_{i-1})=c\sum_{||\Delta_x||\to 0}f(c_i)(x_i-x_{i-1})$$
Suppose $c=0$ then the theorem is trivial. Now if $c \neq 0$, let $\beta= \sum_{j=1}^ncf(c_j)\Delta x_j=c\sum_{j=1}^nf(c_j)\Delta x_j$, where $\Delta x_j=x_j-x_{j-1}$, note that $\beta$ is the Riemann sum of $cf$. Now since $f$ is integrable we now that for every $\varepsilon>0$, $$|\sum_{j=1}^nf(c_j)\Delta x_j-\int_{a}^bfdx|< \frac \varepsilon {|c|}$$ for some partition $P$. Multiplying the inequality by $|c|$ you get $|\sum_{j=1}^ncf(c_j)\Delta x_j-c\int_{a}^bfdx|< \varepsilon$ then $|\beta-c\int_{a}^bfdx|< \varepsilon$ for a partition $P$ this shows that the function $cf$'s Riemann sum converges to $c\int_{a}^bfdx$ , hence $cf$ is Riemann integrable and $\int_{a}^bcfdx=c\int_{a}^bfdx$.