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I was reading the proof of Egoroff's theorem before which Lemma 10 was proved, I was unable to understand the difference between Lemma 10 and Egoroff's theorem but after going through the answer to this question -

Egoroff's theorem in Royden Fitzpatrick (comparison with lemma 10)

I find it little clear, but still I am unable to understand the proof of the lemma 10, Egoroffs theorem statement

Proof of lemma 10

Any help guys!

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    How was $|f_{n} - f|$ measurable , as we cant say for sure that $f$ is measurable , there is alemma which states that $f$ must be measurable on $E - E_{0}$ where $E_{0}$ is the set of zero meaure , but also i think the above statement will hold if $E_{0} = \phi$ . – 2017-02-21
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    Also how we get set $E_{n}$ to be ascending collection? – 2017-02-21

2 Answers 2

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Pay attention, Lemma 10 is not Egoroff's Theorem ! Egoroff's Theorem states that for any $\delta > 0$, there exists $A$ such that $m(E\setminus A)< \delta$ and $f_n \to f$ uniformly in $A$.

In other words: for any $\delta > 0$, there exists $A$ such that $m(E\setminus A)< \delta$ and, for any $\eta > 0$, there exists $N$ such that $|f_n(x)-f(x)|<\eta$, for any $n\geq N$ and for any $x\in A$. Note that the set $A$ depends only on $\delta$.

As you can see Lemma 10, instead, says something similar but with a big difference: the set $A$ depends on $\delta$ and $\eta$ ! In other words, in Lemma 10, for any $\delta$ and $\eta$ you find a set $A$ with certain properties, but if you change $\eta$ the set $A$ may be different. In Egoroff's Theorem you want a set $A$ which depends only on $\delta$.

It should be more precise to use a different notation: in Egoroff's Theorem I would write $A_\delta$, while in Lemma 10 I would write $A_{\delta,\eta}$.

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  1. Lemma 10 is exactly the same statement as Ergodoff, and the proof given above is the proof of the theorem, not the lemma, as nothing is to be prove in it. Indeed, we just throw in the definition of uniform convergence: $f_k \to f$ uniformly on $X$ means for all $\eta > 0$, there exists $N$ such that for all $k \geq N$ and for all $x \in X$, $|f_k(x) - f(x)| < \eta$

  2. The simple limit of measurable functions is a measurable function, such that $f$ is measurable. Moreover, the difference of two measurable functions is measurable, meaning $f_n - f$ is measurable for all $n$. Eventually, the absolute value of a measurable function is measurable, yielding that $|f_n - f|$ is measurable as well. I'll advise you to work out the proof for these statements which are some of the building blocks of the theory.

  3. A collection of sets $\{A_n\}$ is said to be ascending if $A_n \subset A_{n+1}$ for all $n$. It's trivial to see that $\{E_n\}$ is an ascending collection of sets.
  4. $m(E) < \infty$ means $E$ is of finite measure, which is one of the very hypothesis stated at the very beginning of the theorem.