What's the reasoning in a proof that shows equality of Riemann integral and Lebesgue integral, where the proof is started like:
Since $f$ is bounded, then by adding a constant, we can assume that $f\geq 0$
Intuitively it seems that since $-M \leq f \leq M$, then one can "shift" the whole of $f$ upwards so that the lower bound becomes $\geq 0$. However if one adds a constant then this should become
$$0 \leq f +M \leq 2M$$
And then in order to do the proof, wouldn't one need to prove it for $f+M$, rather than $f$.
But if one then considers upper and lower sums and Riemann integral for $f+M$, rather than $f$, then are these equivalent, why?
Maybe their idea is that even if one integrates the constant as well, then "well it's just an integration constant"?
Yes, shifting $f$ up by $M$ amounts to replacing $f$ by $f+M$. So you have a new function for which you need to establish equality of Riemann and Lebesgue integrals. You can rename this new function any way you like, for example you could write $g:=f+M$, and then proceed with the proof using $g$. But most authors just keep using $f$, and impose the additional condition that $f\ge 0$.
Why is it legal to impose this additional condition? After you've proven that the Riemann and Lebesgue integrals coincide in the case that $f$ is bounded and non-negative, the proof for general bounded $f$ follows since $\int (f+M)=(\int f) + M$ for both Riemann and Lebesgue integrals; this last assertion does require proof, though.