# Null Sets and uniform convergence

• May 15th 2010, 02:13 PM
WannaBe
Null Sets and uniform convergence
Prove that if a sequence of continous functions uniformly converges on [a,b] then the union of their graphs is a null set.
In other words:
Prove that if the sequence $f_n:[a,b] \to R$ converges uniformly on [a,b] then the set : $A=(x,f_n) | x \in [a,b],n \in N$ is a null set...

I know the function f is continous ( $f_n \to f$ ) and that the graph of the function f and of each $f_n$ if a null set....
Can't figure out how to prove what I need to prove

• May 15th 2010, 09:34 PM
dwsmith
Quote:

Originally Posted by WannaBe
Prove that if a sequence of continous functions uniformly converges on [a,b] then the union of their graphs is a null set.
In other words:
Prove that if the sequence $f_n:[a,b] \to R$ converges uniformly on [a,b] then the set : $A=(x,f_n) | x \in [a,b],n \in N$ is a null set...

I know the function f is continous ( $f_n \to f$ ) and that the graph of the function f and of each $f_n$ if a null set....
Can't figure out how to prove what I need to prove

$P\rightarrow Q\equiv P\wedge$ ~ $Q$
• May 16th 2010, 03:21 AM
WannaBe
...
Quote:

Originally Posted by dwsmith
$P\rightarrow Q\equiv P\wedge$ ~ $Q$
If the set A isn't a null set, then there is an $\epsilon >0$ such as there is no finite amount of rectangles that cover the set A and their sum is less than $\epsilon$ .
In other words, we assume by contradiction that there is an $\epsilon >0$ such as every finite amount of rectangles that cover the set A- their sum is $\geq \epsilon$ ...