Let $f_1 , f_2: I\mapsto \mathbb{R}$ bounded functions. Show that $L(f_1)+L(f_2)\leq L(f_1+f_2)$ where $L(F)$ is the supremum of the lower sums of the Riemann integral.

I tried to by contradicction $L(f_1)+L(f_2)> L(f_1+f_2)$ and isoleting $L(f_1)> L(f_1+f_2)-L(f_2)$ and usig the definition of sup.

I also tried to use $\sup(A)+\sup(B)=\sup(A+B)$ but it din't work.



For a partition $\Delta=[t_0,t_1,\dots,t_n]$ of $I$, the corresponding lower Riemann sum is $$ L_\Delta(f) = \sum_{i=1}^n (t_i-t_{i-1})f(x_i^*) $$ where $x_i^*\in[t_{i-1}, t_i]$ satisfies $$f(x_i^*)= \inf_{t_{i-1}\leq t\leq t_i}f(t).$$

Since $$\inf_{t_i\leq t\leq t_i} f_1(t) + \inf_{t_i\leq t\leq t_i} f_2(t)$ \leq \inf_{t_i\leq t\leq t_i} f_1(t) + f_2(t),$$ we have for any partition $\Delta$, $$ L_\Delta(f_1) + L_\Delta(f_1) \leq L_\Delta(f_1 + f_2). $$ Taking the supremum over all partitions gives $$ L(f_1) + L(f_2) \leq L(f_1 + f_2). $$


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