If I have 2 Lebesgue Integrable functions $f,g$ defined on the same set A such that: $$ f > g \qquad \hbox{a.e on A}$$ Does this imply that: $$ \int_{A} f d\mu > \int_{A} g d\mu$$I'm not sure that a strict inequality holds here and if it doesn't, does anyone know of any nice results in which I can obtain a strict inequality?
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On
If $A$ has zero measure, then there is no clue: both integrals are $0$. Otherwise consider the descending chain of (measurable) sets indexed by $n \ge 1$ $$A_n = \left\{ t \in A : f(t)-g(t) \le \frac{1}{n} \right\}$$ Then $\bigcap_n A_n$ has measure zero, so $\lim_n \mu(A_n) = 0$. This means that there exists some $n$ such that $\mu(A_n) \le \frac{1}{2}\mu(A)$. Now consider $$\int_A f-g \ge \int_{A \setminus A_n} f-g \ge \int_{A \setminus A_n} \frac{1}{n} = \frac{1}{n}\mu(A \setminus A_n) >0 $$ so, since $f,g$ are Lebesgue integrable (their integrals are finite) you have $\int_A f > \int_A g$.
All of this works if the measure of $A$ is finite. Otherwise, restrict yourself to a subset $A'\subset A$ of finite measure, and use the same argument.
No, it does not necessarily hold. Consiser a $\mu$-measurable set $A$ of measure $0$. Define $f := \chi_A$ und $g := 0$. Then $$ \int_A f \, \mathrm d\mu = \int_B g \, \mathrm d\mu = 0 \; .$$ It holds though, if $A$ has a strict positive measure. Can you proove this?