Let $I\subset \Bbb{R^n}$ be a compact interval and $f:I\to \Bbb{R},g:I\to \Bbb{R}$ bounded functions. Show that $\bar\int_I (f+g) \leq \bar\int_I f + \bar\int_I g$.
Let $M_j(f):=\sup f(I_j)$ and $M_j(g):=\sup g(I_j)$. We have that $M_j(f+g)\leq M_j(f)+M_j(g)$. Since $\bar\int_I (f+g):=\inf\{U((f+g),P)\}.$ We could then write $$\bar\int_I (f+g)\leq U((f+g),P)\leq U(f,P)+U(g,P),$$
which means $\bar\int_I (f+g)$ is a lower bound for $\{U(f,P)+U(g,P)\}$. If we define the infima of this set as $y$, we'd have that $y\geq \bar\int_I (f+g)$. How can I finish this proof?
Since you have only one partition $P$ you couldn't take the infimum. Instead you could try to split the partition as follows. Take two different partitions of $I$, call them $P_1$ and $P_2$ and define $P=P_1 \cup P_2$ which is again a partition of $P$. Then you get the following inequality. \begin{align*} \overline{\int} (f+g) \leq U(f+g, P_1\cup P_2)\leq U(f,P_1\cup P_2)+U(g,P_1\cup P_2) \leq U(f,P_1) + U(g,P_2) \end{align*} Fixing the partition $P_2$ the left take infimum over $P_1$, we get \begin{align*} \overline{\int} (f+g)\leq \overline{\int}f+U(g,P_2) \end{align*} Now take infimum over $P_2$, we get \begin{align*} \overline{\int}(f+g)\leq \overline{\int}f +\overline{\int} g \end{align*} $\textbf{Edit}$:"By fixing partition $P_1$ taking infimum over $P_2$" I meant the following, \begin{align*} \inf_{P_2:P_2\text{ a partition of $I$}} \biggl( U(f,P_1)+U(g,P_2)\biggr) = U(f,P_1)+ \inf_{P_2:P_2\text{ a partition of $I$}}U(g,P_2) \end{align*} The above is legitimate because $P_2$ is not involved in the term $U(f,P_1)$