If $\sup(f+g)\le \sup \ f + \sup \ g$, then why $\underline\int_{A}f(x)\ dx + \underline\int_{A}g(x)\ dx \le \underline\int_{A}[f(x) + g(x)]\ dx$

Asked Oct 13 '17 at 19:05

Active Oct 13 '17 at 19:12

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If $\sup(f+g)\le \sup \ f + \sup \ g$, then why

$$\underline\int_{A}f(x)\ dx + \underline\int_{A}g(x)\ dx \le \underline\int_{A}[f(x) + g(x)]\ dx$$

What's wrong?

edited Oct 13 '17 at 19:12

puzzleweaver

asked Oct 13 '17 at 19:05

Poperton

Where is the disagreement with the lower integral inequality which is correct, and a property of the supremum? Why would you not provide feedback or ask for clarification in the previously posted question/answer? – RRL Oct 13 '17 at 19:37
@RRL because the proof he gave seems correct. The disagreement is that in one side, the sup is greater than the sum of the sups, while in the other side, not – Poperton Oct 13 '17 at 20:23

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