Following my previous question about Riemann-Stieltjes integration, I'm asking this problem.
Let $f_1$ and $f_2$ be bounded on $[a,b]$ and $\alpha$ is increasing on $[a,b]$. Define $f = f_1 +f_2$. Either prove the statements or give a counterexample to show that the given statements are false:
$1.$ $U(P,f,\alpha) \le U(P,f_1,\alpha) + U(P,f_2,\alpha)$
$2.$ $L(P,f,\alpha) \le L(P,f_1,\alpha) + L(P,f_2,\alpha)$
My try: The first statement is correct. We know that $$U(P,f,\alpha) = \sum_{i =1}^{n} M_i \Delta \alpha_i , \ M_i = \sup\{f(x):x\in [x_{i-1}, x_i]\} \\ U(P,f_1,\alpha) = \sum_{i =1}^{n} N_i \Delta \alpha_i , \ N_i = \sup\{f_1(x):x\in [x_{i-1}, x_i]\} \\ U(P,f_2,\alpha) = \sum_{i =1}^{n} Q_i \Delta \alpha_i , \ Q_i = \sup\{f_2(x):x\in [x_{i-1}, x_i]\}$$ So we have $$U(P,f_1,\alpha) + U(P,f_2,\alpha) = \sum_{i =1}^{n} N_i \Delta \alpha_i + \sum_{i =1}^{n} Q_i \Delta \alpha_i = \sum_{i =1}^{n} (N_i+Q_i) \Delta \alpha_i \ge \sum_{i =1}^{n} M_i \Delta \alpha_i = U(P,f,\alpha)$$ The inequality comes from the fact $$\sup(A+B)\le \sup(A) + \sup(B)$$ The second one is false. Let $f_1 = x$ and $f_2 = -x$ and $a = 0,b=2, P = \{0,1,2\}$. Therefore $$L(P,f_1,\alpha) = 1,L(P,f_1,\alpha) = -3 \\ L(P,f,\alpha) = 0$$ Obviously $0 \not \le -2$ and this is false.
Is my answer correct? If not please explain why.
Your counterexample for case (2) is correct.
More generally for any interval $I$ we have for all $x \in I$,
$$\inf_{x \in I} f_1(x) + \inf_{x \in I} f_2(x) \leqslant f_1(x) + f_2(x) = f(x),$$
which implies
$$\inf_{x \in I} f_1(x) + \inf_{x \in I} f_2(x) \leqslant \inf_{x \in I}f(x), \\L(P,f_1,\alpha) + L(P,f_2,\alpha) \leqslant L(P,f,\alpha)$$
Together the inequalities
$$L(P,f_1,\alpha) + L(P,f_2,\alpha) \leqslant L(P,f,\alpha) \leqslant U(P,f,\alpha)\leqslant U(P,f_1,\alpha) + U(P,f_2,\alpha)$$
can be used to prove that if $f_1$ and $f_2$ are Riemann-Stieltjes integrable, then so is $f = f_1+f_2$. The converse is not true and this is connected to the fact that the reverse of the LHS inequality is false.