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David Roberts ♦
David Roberts ♦
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The usual Fubini's theorem(see the Wikipedia articleWikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?

I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.

The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.

The usual Fubini's theorem(see the Wikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?

I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.

The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.

The usual Fubini's theorem(see the Wikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?

I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.

The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.

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Makoto Kato
Makoto Kato
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Fubini's theorem without completeness or $\sigma$-finiteness conditions

The usual Fubini's theorem(see the Wikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?

I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.

The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.