Let $f,g:[a,b]\rightarrow \mathbb{R}$ be bounded functions which are Riemann integrable. I want to show $$L(f,[a,b])+L(g,[a,b])\leq L(f+g,[a,b]).$$
Now I know that if a function is Riemann integrable, its lower Riemann integral equals its upper Riemann integral. So I did the following:
\begin{equation} \begin{split} & L(f,[a,b]+L(g,[a,b]) \\[6pt] = {} & U(f,[a,b])+L(g,[a,b])\\[6pt] = {} & \lim_{n\to\infty} \left(\left(\frac{b-a}{n}\right) \sum_{j=1}^n \sup(f)_{[x_{j-1},x_j]}\right)+ \lim_{n\to\infty} \left(\left(\frac{b-a}{n}\right)\sum_{j=1}^n \inf(g)_{[x_{j-1}, x_j]} \right)\\[6pt] ={} & \lim_{n\to\infty} \left(\frac{b-a}{n} \left(\sum_{j=1}^n \sup(f)_{[x_{j-1},x_j]} + \inf(g)_{[x_{j-1},x_j]}\right)\right) \end{split} \end{equation} On the other hand, we have:
\begin{equation} \begin{split} L(f+g,[a,b])&=\lim_{n\to\infty} \left(\frac{b-a}{n} \left( \sum_{j=1}^n \inf(f+g)_{[x_{j-1},x_{j}]}\right)\right)\\[6pt] &= \lim_{n\to\infty} \left(\frac{b-a}{n} \sum_{j=1}^n \left( \inf(f)_{[x_{j-1}, x_{j}]}+\inf(g)_{[x_{j-1},x_j\}} \right) \right)\\ \end{split} \end{equation}
I do not know how the first equation can be less than or equal to the second. I would think that since you are adding up a bunch of supremums of f in the first, that that would actually be greater than the second because you are only adding up infimums. What am I doing wrong here?