Let $f:(\mathbb{R},\mathbb{L(\mathbb{R})},\lambda )\rightarrow[0,+\infty) $ Lebesgue-measurable and where $\mathbb{L(\mathbb{R})}$ is the Lebesgue space (I'm sorry i didnt know how to denote it with latex).Show that $\int_{[x,+\infty]}^{}f\,d\lambda\to 0 \,\,\, ,as\,\,\, x\to +\infty$.
I'm not sure my solotions makes sense and I'm insecure about using {${+\infty}$} as a singleton. Is my logic correct ? Is there another way to solve this ? Here goes my approach:
It is known that for every non-negative measurable $f$ exists a sequence of simple and non-negative functions $s_n$ such as : $s_n\uparrow f$.
Then it follows that if we take the restriction of $f$ at $[x,+\infty]$ : $f|_{[x,+\infty]}$ there exists $s_n:[x,+\infty]\rightarrow [0,+\infty]$ of simple and non negative functions so that :$$\int_{[x,+\infty]}^{}f\,d\lambda=lim_n\int_{}^{} s_n\,\,\,d\lambda$$
Then as $x\to+\infty$ and $[x,+\infty]=\bigsqcup_{i=1}^{m}A_i^{(n)} $ we get $lim_n\int s_n d\lambda=lim_n\int\sum_{i=1}^{m}a^{(n)}_i*1^{(n)}_{A_i}=lim_n\sum_{i=1}^{m}a^{(n)}_i\lambda(A_i)^{(n)}=lim_na_i^{(n)}\lambda({+\infty})=0$ because {${+\infty}$} is a singleton.
Disclaimer $^{(n)}$ is not a power ,it is an index.
Thank you in advance !