I need to prove: $(\Bbb{X}, \Sigma, \mu)$ is a measure space and let $f \in L^1(\mu)$.
Then, $\displaystyle \lim_{t\to \infty} \mu(\{ x : |f(x)| \gt t\}) = 0$
My thoughts: We proved in class that if $f \in L^+$ and $\int_{\Bbb{X}}f\ d\mu \lt \infty$ then $\mu(\{ x : f(x) = \infty\}) = 0$ and I think I have to incorporate it somehow into my proof. Am I on the right track?
Proving the above is equivalent to showing that - $ \lim_{n\to\infty} \mu \{ x : |f(x)|>n \} =0$.
Define the set $ K_n = \{x \in X : |f(x)| \leq n\}$ Then, since $f \in L^1(\mu)$, we have that $ \lim_{n\to\infty} \int_{K_n} |f| d\mu = \int_{X} |f|d\mu$ by the Lebesgue Dominated Convergence theorem. Hence, the above equation becomes, $ \lim_{n\to\infty} \int_{{K_n}^c} |f| d\mu= 0$ Since, $ 0\leq n\mu(K_n)^c \leq \int_{{K_n}^c} |f|d\mu$ Hence taking limits on both sides and by the the Sandwich theorem, we get the required result.