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I have already solved part (a), anyway here is resolved by Two functions agreeing except on set of measure zero, and for an answer to both questions look here Integration over a bounded set, the fact is that I am not very familiar with theory of measurement, I only know the basics, so I would like to know how I show (b), in an easy way using the basic taboo of the measure, I know that as $f(x)\leq g(x)$ for $x\in S$, then I just have to prove that $D:=\{x\in S:f(x)<g(x)\}$ has zero measure, how do I do this? Thank you very much.

edm
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user424241
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1 Answers1

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If $\{x \in S : f(x) < g(x)\} = \cup_{n \ge 1} \{x \in S : f(x) < g(x)-\frac{1}{n}\}$ has positive measure then there must be some $n \ge 1$ for which $E_n := \{x \in S : f(x) < g(x)-\frac{1}{n}\}$ has positive measure, but then $\int_S f = \int_{E_n} f + \int_{S\setminus E_n} f \le \int_{E_n} g-\frac{1}{n} + \int_{S\setminus E_n} g = \int_S g - \frac{1}{n}\mu(E_n) < \int_S g$, contradiction.

mathworker21
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