Consider $E\in\mathscr{F}$(sigma-algebra), and $E\in\Omega$ defined on a measure space $(\Omega,\mathscr{F},\mu)$. Suppose $\mu(E)<\infty$, and $\{f_n\}$ is a sequence of measurable functions on $E\to\mathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:E\to\mathbb{R}$ which is also finite almost everywhere. Then $f_n\to f$ almost uniformly in $E$.

Proof: By omitting a subset of $E$ of zero measure, we may assume that all the functions $f_n$ and $f$ are finite that $f_n(x)\to f(x)$ for all $x\in E$.

For positive integers $m,n$ let $$A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\}.$$

Then for fixed $m$, the sequence $A^m_1, A^m_2, \dots,$ is an increasing sequence of measurable sets converging to $E$. Since $\mu(E)$ is finite, by theorem 3.2 there is a positive integer $N_m=N_m(m)$ such that $$\mu(E-A^m_n)<\frac{\epsilon}{2^m}$$ for $n \geqslant N_m$. If we put $$F_\epsilon=\bigcup_\limits{m=1}^{\infty}(E-A^m_{Nm}),$$ then $\mu(F_\epsilon)<\epsilon$. Furthermore, given $\delta>0$, we can choose $m$ so that $\frac{1}{m}<\delta$ and then $|f_i(x)-f(x)|<\delta$ for all $i\geqslant N_m$ $x\in (E-F_\epsilon)$, so that $f_n\to f$ uniformly on $(E-F_\epsilon)$$\blacksquare$ .

I have made a question about this theorem on another post. Another doubt troubles me. How can we go from a situation in which we have convergence a.e as theorem states to uniform convergence? Where is that explicit in the proof? What I understand of the proof is that on the set $E-F_\epsilon$ the function converges, because it converged as the theorem states. Thanks in advance!


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