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Let be a null measure set, we have to prove that is also a null measure set.
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Let us show that is a null measure set.
Therefore and thus is a null measure set.
Since is absolutely continuous on the set is a null measure set for each.
Let us show that the set is a null measure set.
Since contains no positive measure subset we can ensure that is a null measure set for all,, and since increases with and, we conclude that is a null measure set.
Since and are absolutely continuous there is a null measure set such that for all, thus for we have (3.8).
Analogous arguments show that is a null measure set, thus the proof of Claim 3 is complete.
Suppose that there is a null-measure set such that conditions and in Theorem 3.1 hold for and assume moreover that the following condition holds: there exists such that for all,, one has.
Suppose that there is a null-measure set such that conditions and in Theorem 3.1 hold for and assume moreover that the following condition holds: for every there exists such that for all,, and one has.
For every n ∈ N, the set { t ∈ I n : x ( t ) = γ n ( t ) } is a null-measure set, for otherwise we would have in a positive measure set | γ n ″ ( t ) | = | x ″ ( t ) | ≤ M ( t ) (because x ∈ K ), a contradiction with (C2)∗.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com