Riemann integral property proof

Question

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

Answer 1

For any Riemann function (i.e., for any partition of $\;[a,b]\;$):

$$\sum_{||\Delta_x||\to 0}(cf)(c_i)(x_i-x_{i-1})=c\sum_{||\Delta_x||\to 0}f(c_i)(x_i-x_{i-1})$$

and thus the same as above is true for the supremum (infimum) of all the sums (and the same if you're used to lower/upper sums or Darboux sums).

Answer 2

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 $\int_{a}^bcfdx=c\int_{a}^bfdx$.