Proof question about the definition of the integral via Darboux sums

Question

I am having trouble on how to proceed with this question and would appreciate some help.

$a,b,c: [f,g] \rightarrow \mathbb{R}$ are Riemann integrable, and $a(z) ≤ b(z) + c(z)$ for all z ∈ [f,g]. Using the definition of the integral and Darboux sums, show that $\int_{[f,g]} a \leq \int_{[f,g]} b + \int_{[f,g]}c$

Right now, I have that $a(z) ≤ b(z) + c(z)$ gives $\sup(a) \leq \sup(b+c) \leq \sup(b)+\sup(c)$

I am not sure how to proceed after this.

Thanks!

Answer 1

We know that if $a(z)<d(z)$ and $a(z),d(z) \in R[f,g]$, then $\int_f^g a(z)dz < \int_f^g d(z)$.

this is because $\forall \epsilon>0, \exists \delta>0\text{ s.t }$ any tagged partition $\|\dot{P}\|<\delta \Rightarrow \int_f^ga(z)-\epsilon<R(a(z),\dot{P})<R(b(z),\dot{P})<\int_f^gd(z)+\epsilon$ .

Now $b(z),c(z) \in R(f,g) \Rightarrow b(z)+c(z) \in R[f,g]$.

combining with above fact gives the result.