# Thread: Null Sets and uniform convergence

1. ## 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 $\displaystyle f_n:[a,b] \to R$ converges uniformly on [a,b] then the set : $\displaystyle A=(x,f_n) | x \in [a,b],n \in N$ is a null set...

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

2. 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 $\displaystyle f_n:[a,b] \to R$ converges uniformly on [a,b] then the set : $\displaystyle A=(x,f_n) | x \in [a,b],n \in N$ is a null set...

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

$\displaystyle P\rightarrow Q\equiv P\wedge$ ~$\displaystyle Q$

3. ## ...

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