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In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf

Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below).

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.

In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf

Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below).

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.

In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf

Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below).

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.

I would supplement Joel David's fine answer by pointing out that, before anyone draws hasty philosophical conclusions from this consistency fact, the failure of countable additivity of the Lebesgue measure (taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.