Let $f$ be $f: [a,b] \to \mathbb{R}$. For a partition $P$ of the interval $[a,b]$ we define: $L_{f}(P) = \sum_{i=1}^{n} m_{i}\triangle x_{i}$ with $m_{i} = min \{ f(x), for f(x) \in [x_i-1,x_i] \}$ (We call this the lower sum of $f$ with partition $P$). The upper sum is defined in a similar way, you just have to look at the maximum values of the function instead of the minimum values in each interval.
So, my question is: how do I proof formally that $L_{\alpha f}(P) = \alpha U_{f}(P)$ for $\alpha > 0$? I can see why this little fact is true when I draw some functions or when I calculate is, but I can't find a formal proof.
Ps: I would start the proof with the mention that the uppersum is greater than or equal to the lowersum of the function in this case (more general: regardless of the chosen partition). We can multiply both by $alpha$ and maintain the inequality. But what now?