Prove (using Riemann's definitions) that if$ f_1,f_2,...,f_n$ are integrable on$[a,b]$ and $c_1,c_2,...c_n$ are constant then $c_1f_1+c_2f_2 + ...+c_nf_n $is integrable: $$\int_{a}^{b} (c_1f_1+c_2f_2 +...c_nf_n) = c_1 \int_a^b f_1(x) dx + c_2 \int_a^b f_2(x) dx +...+ c_n\int_a^b f_n(x) dx$$
Let's have as assumption that it has already been proven that:
- 1: "If $f$ and $g$ are integrable on $[a,b]$, then so is $f+g$"
- 2: "If $f$ is integrable on $[a,b]$ and $c$ is a constant then $cf$ is integrable on $[a,b]$"
As we know that $f_1, f_2, ... f_n$ are integrable $\forall \epsilon >0$ there is a $\delta > 0$ such that $$|\sigma_f- L_1 | <\epsilon$$ and $$||P||<\delta$$ Where the limit L when $||P|| \rightarrow 0$ or $n \rightarrow \infty$ $$(b-a)\lim\limits_{ n \rightarrow \infty} \sum_{j=1}^{n} f_j(x) = L_1$$
We know as given that $cf$ is integrable, let's check if it is then so for $c_1f_1 + c_2f_2 + ... + c_nf_n$ The Rieman sum $$\sigma_{cf}= \sum_{j=1}^{n} c_jf_j(x) (x_j-x_{j-1})$$ $$\sigma_{cf}= c(b-a) \sum_{j=1}^{n} f_j(x)$$ So the limit $L_2$ $$c(b-a)\lim\limits_{ n \rightarrow \infty} \sum_{j=1}^{n} f_j(x) = L_2$$
We know that $\sigma_{cf} - L_2 = c(\sigma_f - L_1)
$$\left|{\sigma_{cf} - L_2}\right| < \epsilon_2$$ $$\left|\frac{\sigma_{cf} - L_2}{c}\right| < \frac{\epsilon_2}{|c|}$$ $$\left|{\sigma_{c} - L_1}\right| < \frac{\epsilon_2}{|c|} = \epsilon_1$$
It follows that $c_1f_1 + c_2f_2+ ...+c_nf_n$ is integrable
I am not too certain if my "Riemann" argumentation is properly done here. any input or help is greatly appreciated.
Approach A.
With (1) and (2) you can use induction.
Approach B.
Without (1) and (2) consider for a subinterval $I$ of a partition,
$$\sup_{x,y \in I}\left|\sum_{j=1}^n c_j f_j(x) - \sum_{j=1}^n c_j f_j(y)\right| \leqslant \sup_{x,y \in I}\sum_{j=1}^n |c_j| |f_j(x) - f_j(y)| \\ \leqslant \sum_{j=1}^n |c_j|( \sup_{x,y \in I}|f_j(x) - f_j(y)| )$$
Recall that
$$U(P,f_j) - L(P,f_j) = \sum_{k=1}^n\sup_{x,y \in [x_{k-1},x_k]}|f_j(x) - f_j(y)| (x_k - x_{k-1}) $$
Now what can you say about $U(P, \sum_j c_jf) - L(P, \sum_j c_jf_j)$?
Approach C.
If you must use Riemann sums, note that
$$S(P, \sum_j c_j f_j) = \sum_{k=1}^m \sum_{j=1}^n c_j f_j(\xi_k)(x_k - x_{k-1})= \sum_{j=1}^n c_j\sum_{k=1}^mf_j(\xi_k)(x_k - x_{k-1})\\ = \sum_{j=1}^n c_j S(P, f_j)$$
and
$$\left|S(P, \sum_jc_j f_j) - \sum_{j=1}^n c_j \int_a^b f_j\right| \leqslant \sum_{j=1}^n |c_j| \,\left|\, S(P, f_j) - \int_a^b f_j\right| $$