Riemann integral $\underline{\int_a^b} \alpha f = \alpha \underline{\int_a^b} f$

Question

During our introduction to Riemann integrals and Darboux sums we recieved we encountered this theorem:

Let $f$ be a bounded function in $[a,b]$, Then if $\alpha >0$ then $\underline{\int_a^b} \alpha f = \alpha \underline{\int_a^b} f$.

So far I've proven by contradiction that $$\alpha \inf f = inf (\alpha f)$$.

I've used this fact to show that for some partition $P$: $$L(\alpha f, P)= \alpha L(f,P)$$

But I'm not sure how to translate my findings to arguments regarding lower integarls. $$ \underline{\int_a^b} \alpha f = \sup \,L( \alpha f,P) $$ and $$ \alpha \underline{\int_a^b} = \alpha \sup \,L(f,P) $$

Answer 1

For each partition $P$, we have$$L(\alpha f,P)=\alpha L(f,P)$$and therefore\begin{align}\underline{\int_a^b}\alpha f&=\sup_PL(\alpha f,P)\\&=\sup_P\alpha L(f,P)\\&=\alpha\sup_PL(f,P)\\&=\alpha\underline{\int_a^b}f.\end{align}

Answer 2

You can take out $\alpha$ because of the following results.

Let $\{x_i\}_{i \in I}$ be a bounded collection of real numbers. Then for every $\alpha \in [0,\infty)$ it holds the following: $$\sup_I{\alpha x_i} = \alpha \sup_I{x_i}$$ Proof: $$\alpha x_i \leq \alpha \sup_I {x_i}$$ for every $i$ so we can pass to the supremum and we get $$\sup_I{\alpha x_i} \leq \alpha \sup_I{x_i}$$ For the converse inequality let $i^*$ be such that $\sup_I{x_i} - x_{i^*} < \frac{\epsilon}{\alpha}$ Now: $$\alpha \sup_I {x_i} \leq \epsilon + \alpha x_{i^*} \leq \epsilon + \sup_I {\alpha x_i}$$ Since $\epsilon$ is arbitrary we get: $$\alpha \sup_I {x_i} \leq \sup_I {\alpha x_i}$$