Show that $\mu(\{x:f(x)>0\}) > 0 \implies \mu(\{x:f(x)>1/N\}) > 0$

Question

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I'm not sure how to do this problem. I first notice that

$$\{\omega\in\Omega:f(\omega)>1/N\} \uparrow \{\omega\in\Omega:f(\omega)>0\}$$

and therefore

$$\mu(\{\omega\in\Omega:f(\omega)>1/N\}) \uparrow \mu(\{\omega\in\Omega:f(\omega)>0\})$$

but I'm not sure if this is helpful.

Here is my argument:

$$\mu(\{\omega\in\Omega:f(\omega)>0\})>0 \implies \exists \epsilon>0 \;\;st\;\; \mu(\{\omega\in\Omega:f(\omega)>\epsilon\})>0$$

Let $N$ be large enough such that $1/N < \epsilon$.

Is this enough? Does it even make sense?

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EDIT:

Let $A_n, n\in \mathbb{N}$ be such that $\mu(A_n)=0$. Then

$$\mu\left(\bigcup_n A_n\right)\leq \sum_n \mu(A_n) = 0$$

Using this result

$$\{\omega\in\Omega:f(\omega)>0\} = \bigcup_N \{\omega\in\Omega:f(\omega)>1/N\} $$

Leading to a contradiction if all the sets have measure zero.

Answer 1

Hint: A countable union of sets with measure $0$ has measure $0$ (show this!). How can you apply this here?

Answer 2

Your EDIT uses countably additive property, and it goes through.

Your first solution uses the monotonic of sequence, and it actually goes through, as you know that $\lim_{n}a_{n}=L>0$, then for sufficiently large $n$, $a_{n}>0$.