This is a follow-up to the question here with, hopefully, the correct inequalities.
I am trying to check if these following inequalities are true
$$ \sup_{0 < s < 1} \left| (f(s) + g(s)) \right| \leq \sup_{0 < s < 1} \left| f(s) \right| + \sup_{0 < s < 1} \left| g(s) \right| $$
and $$ \sup_{0 < s < 1} \left| \int_0^s f(u) \ d \mu(u) \right| \leq \sup_{0 < s < 1/2}\left| \int_0^s f(u) \ d \mu(u) \right| + \sup_{1/2 < s < 1} \left| \int_{1/2}^s f(u) \ d \mu(u) \right|. $$
where $\mu$ is the Lebesgue measure. Unlike my previous question here, the absolute values are inside the sup.
I think both inequalities are true in general. I argue about them as follows: assume that the maximum is at $s^*$, then
$$ \sup_{0 < s < 1} \left| (f(s) + g(s)) \right| = \left| f(s^*) + g(s^*) \right| \leq \left| f(s^*) \right| + \left| g(s^*) \right| \leq \sup_{0 < s < 1} \left| f(s) \right| + \sup_{0 < s < 1} \left| g({s}) \right| $$
I can also argue for the second inequality using a similar idea by considering the two cases: when $s^*$ is in $[0,1/2]$ and when it's in $[1/2,1]$. However, I don't think this proof is correct in general, since such an $s^*$ might not exist.
How would I go about proving these inequalities in general?
Also, is the inequality with $\mu$ true for other measures as well?
Your argument is wrong because the two supremums may be attained at different points. A correct proof is as follows: $|f(t)+g(t)| \leq |f(t)|+|g(t)| \leq \sup \{|f(s)|: 0<s<1\}+\sup \{|g(s)|: 0<s<1\} $. This is true for all $t$. Take sup over $t$ to get the first inequality. The second one is similar since the integral over $(0,s)$ splits into integral over $(0,1/2)$ and $(1/2,s)$ is $s >1/2$ (and you can bound by just the first term on the right side when $s<1/2$.