Are all real-valued continous functions in $\mathbb{R}^2$ on a compact space integrable?

Question

If you look at all continous functions on a compact intervall in $\mathbb{R}$ this statement is true. They are all integrable (proof). Is this true in $\mathbb{R^2}$ and higher dimensional spaces?

I can't think of a function where this is false but I'm not quite sure. I tried to apply Fubini's theorem to break it down to but therefore the functions needs to be integrable in the first place.

Is there an simple argument I missed?

It's not enough to know that they're continuous and bounded? — Randall, Dec 04 '17 at 15:07

Jason · Accepted Answer · 2017-12-04T20:02:05.457

4

Continuous functions on a compact set are bounded, and compact sets have finite Lebesgue measure, thus all such functions are integrable.

edited Dec 04 '17 at 20:02

answered Dec 04 '17 at 15:07

Jason

15,438

+1. But perhaps, for the sake of completeness, it would be good to recall that compact sets in Euclidean space have finite measure. – Aloizio Macedo Dec 04 '17 at 16:32
Fair enough, will edit. – Jason Dec 04 '17 at 20:01

Are all real-valued continous functions in $\mathbb{R}^2$ on a compact space integrable?

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