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Let $f$ and $g$ be Riemann Integrable. Prove $f+g$ is Riemann Integrable and
$$\int_a^bf(x)+g(x)dx= \int_a^b f(x)dx + \int_a^bg(x)dx$$

I have the following:
For a partition $P$ of $[a,b]$
$L(p;f+g)=$
$$\sum_{i=1}^{n} \min(f+g)\Delta x_i$$
$$\ge \sum_{i=1}^{n}\min(f)\Delta x_i + \sum_{i=1}^{n}\min(g)\Delta x_i$$
$$=L(p;f)+L(p;g)$$
Which then implies the equalities of the summations of upper and lower Darboux Integrals.
The only part I need help with is how to show the inequality with the summations.

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Clearly $$f(x)+g(x)\leq f(x)+g(x)$$ for all $x\in[a_i,b_i]$. Then, from the definitions, $$\inf_{x\in[a_i,b_i]}f(x)+\inf_{x\in[a_i,b_i]}g(x)\leq\inf_{x\in[a_i,b_i]}(f(x))+g(x)\leq f(x)+g(x).$$This shows $\inf_{x\in[a_i,b_i]}f(x)+\inf_{x\in[a_i,b_i]}g(x)$ is a lower bound for $f(x)+g(x)$, so the infimum cannot be less than that since the infimum is, by definition, the greatest lower bound. Thus, $$\inf_{x\in[a_i,b_i]}f(x)+\inf_{x\in[a_i,b_i]}g(x)\leq\inf_{x\in[a_i,b_i]}(f(x)+g(x)).$$

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  • $\begingroup$ why is the first inequality you wrote not strictly and equality? $\endgroup$ – user72195 Jul 19 '13 at 14:53
  • $\begingroup$ It is; I just want to emphasize the later inequalities. In fact, that is the reason it is 'clear' :) $\endgroup$ – Clayton Jul 19 '13 at 14:54
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Assume $$\inf_{x\in[a_i,b_i]}(f(x)+g(x))<\inf_{x\in[a_i,b_i]}f(x)+\inf_{x\in[a_i,b_i]}g(x)).$$ Then for some $x_0\in[a_i,b_i]$ $$f(x_0)+g(x_0)<\inf_{x\in[a_i,b_i]}f(x)+\inf_{x\in[a_i,b_i]}g(x)),$$ which implies that $$f(x_0)<\inf_{x\in[a_i,b_i]}f(x))\qquad\text{or}\qquad g(x_0)<\inf_{x\in[a_i,b_i]}g(x)),$$ but both are impossible.

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