In measure theory I encountered Egorov's theorem which states that if ( X , S...

Lucia Grimes

Lucia Grimes

Answered

2022-07-02

In measure theory I encountered Egorov's theorem which states that if ( X , S , μ ) is a measure space such that μ ( X ) < i.e. μ is a finite measure.If ( f n ) be a sequence of measurable functions on X converging pointwise to f : X R ,then f n is almost uniformly convergent to f.
Now in the book the definition of almost uniform convergence is the following:
f n f almost uniformly if for each ϵ > 0,there exists E ϵ X such that μ ( E ϵ ) < ϵ and f n f uniformly on X E ϵ .
Now in some books I have seen that almost everywhere means outside a measure 0 set.
So the definition of almost uniformly convergent should have been f n f almost uniformly if E X such that μ ( E ) = 0 and f n f uniformly on X E.
But unfortunately the definition is not so.In fact the latter condition is stronger.I want to know why the former is taken as a definition and what the problem with the latter one is.I want to understand where I am making mistake in understanding the word "almost".

Answer & Explanation

Sanaa Hinton

Sanaa Hinton

Expert

2022-07-03Added 15 answers

Your alternate definition wouldn't work.
Here is a counterexample: u n ( x ) = x n on [0,1], which converges pointwise to 0 if x < 1 and 1 otherwise. This sequence doesn't converge uniformly on [0,1], nor on the half-open set [0,1), and you can't find a set S of measure 0 such that convergence is uniform on [ 0 , 1 ] S: you would still need values of x arbitrarily close to 1.
However, u n converges uniformly on every compact interval that doesn't contain 1, so you can find S m = ( 1 1 / m , 1 ] such that μ ( S m ) 0, and u n converges uniformly on [ 0 , 1 ] S m for all m.

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