# Measure Theory-Integration Over Non-Lebesgue Measures

Stupid Question: If I'm given a non-euclidean space, such as $S^n$ (n-dimensional unit sphere), with the the uniform measure $\mu$ on it.

I'm also given a function $f:S^n \to \mathbb{R}$ that is bounded, by a constant $C$ . As an analogous to the Lebesgue measure case, can I say that: $\int_{S^n } f d\mu \leq C \mu ( S^n) = C$?

Hope you'll be able to help

Thanks !

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Yes. This follows from the fact that the integral of a non-negative function is non-negative and is true for any measure on any measure space whatsoever. – Qiaochu Yuan Jul 22 '12 at 18:58
Great! THanks a lot ! – joshua Jul 22 '12 at 19:13

Yes. This follows from the fact that the integral of a non-negative function is non-negative and is true for any measure on any measure space whatsoever.

Specifically, we have $C\ge f$ everywhere, so $C-f\ge 0$, and by the above observation

$$\int C-f d\mu \ge 0 \Rightarrow C\mu(S^n) = \int C\ d\mu \ge \int f d \mu.$$

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Another way to see this is that it follows directly from the definition of the integral of an arbitrary nonnegative measurable function: $\int f\,d\mu$ is the supremum of the integrals of simple functions that are bounded by $f$. Each of those simple functions is also bounded by $C$, so its integral is bounded by $C\mu(S^n)$. (If $f$ has a nontrivial negative part, removing it only raises the integral, so we can assume $f$ is nonnegative.)

Of course this assumes $\mu(S^n)$ is finite, but so does the question. If the measure is infinite then so is $\int C\,d\mu$ and the inequality is trivial.

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