- #1

MathematicalPhysicist

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Let ##(X,\mathcal{A},\mu)## be a measurable space, ##E## is a measurable subset of ##X## and ##f## is a measurable bounded function which has a bounded support in ##E##.

Prove that: if ##f\ge 0## almost everywhere in ##E##, then for each measurable subset ##F\subset E##, we have:

$$0\le \int_F fd\mu \le \int_E fd\mu$$

Now, for the proof of the claim: they write in the book that ##f\ge 0## in ##E##, thus by the additive and non-negativity properties of the integral we get: $$\int_E f d\mu =\int_F fd\mu + \int_{E-F} f d\mu \ge \int_F f d\mu + 0 =\int_F fd\mu \ge 0$$

Now I believe in the rhs they used the "fact" that ##f\ge 0## also in ##F##, but the assumption in the claim is that ##f\ge 0## a.e in E, i.e for ##\exists F': F'\subset E## which has zero measure thus ##\int_F' f d\mu = 0##.

I believe this last statement is missing in the proof, i.e that also for zero measured subset of ##E## where ##f<0## the integral is zero.

Other than this little qualm everything seems right.

Am I right?