Re: Egoroff's Theorems Proof

1) We write the set on which $\displaystyle f_n$ converges pointwise to f as an union of such sets. We take $\displaystyle \frac 1m$ instead of $\displaystyle \delta\in\mathbb R_+^*$ in order to get a countable family.

2) Since $\displaystyle \lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\displaystyle \delta=\varepsilon 2^{-m}$ to choose $\displaystyle n_m$ such that $\displaystyle m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $\displaystyle m(E_{m,k})\leq \varepsilon 2^{-m}$ for $\displaystyle k\geq n_m$, but it's not necessary here).

3) We used the fact that $\displaystyle m(A)<\infty$ when we wrote $\displaystyle m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.

Re: Egoroff's Theorems Proof

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Originally Posted by

**girdav** 1) We write the set on which $\displaystyle f_n$ converges pointwise to f as an union of such sets. We take $\displaystyle \frac 1m$ instead of $\displaystyle \delta\in\mathbb R_+^*$ in order to get a countable family.

2) Since $\displaystyle \lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\displaystyle \delta=\varepsilon 2^{-m}$ to choose $\displaystyle n_m$ such that $\displaystyle m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $\displaystyle m(E_{m,k})\leq \varepsilon 2^{-m}$ for $\displaystyle k\geq n_m$, but it's not necessary here).

3) We used the fact that m(A)<\infty when we wrote $\displaystyle m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.

Can the set on which converges pointwise to f be written as $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} ?

Re: Egoroff's Theorems Proof

Yes, you can see that using the definition of the limit.

Re: Egoroff's Theorems Proof

Quote:

Originally Posted by

**girdav** Yes, you can see that using the definition of the limit.

Then what is difference between $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} AND

$\displaystyle \bigcap_{m\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$}

both also refer to pointwise convergence? or one of them is refer to uniform convergence?