$\textbf{The Problem:}$ Let $f$ be a measurable function on $\mathbb R$ with respect to the Lebesgue measure $m$.
$\textbf{a)}$ Suppose that $$m(\{\vert f\vert>\lambda\})\leq(1+\lambda)^{-1}$$ holds for every $\lambda>0.$ For which $p\in\{1,\infty\}$ do we have $f\in L^{p}(\mathbb R)$?
$\textbf{b)}$ Now suppose that $$m(\{\vert f\vert>\lambda\})\leq(1+\lambda)^{-10}.$$ For which $p\in[1,\infty]$ do we have $f\in L^{p}(\mathbb R)?$
$\textbf{My Thoughts:}$
$\textbf{a)}$ It doesn't necessarily follow that $f\in L^1(\mathbb R)$ or $f\in L^{\infty}(\mathbb R)$ since a counterexample in both cases is given by $f(x)=(2x)^{-1}\cdot\mathbf1_{x>0}.$
$\textbf{b)}$ The following formula for the $L^p-$norm will be used $$\|f\|_{p}^{p}=\int_{0}^{\infty}p\lambda^{p-1}m(\{|f|>\lambda\})d\lambda.$$ With that in mind we have the following by hypothesis $$\|f\|_{p}^{p}\leq\int_{0}^{\infty}\frac{p\lambda^{p-1}}{(1+\lambda)^{10}}d\lambda,$$ so if $p-1\leq8$ we are golden, and the integral converges, and so $f\in L^{p}(\mathbb R)$ for all $1\leq p\leq9.$ If $p>9$, then the bound on the $L^p-$norm seems useless. So I think the claim may not follow, so I tried building up counterexamples from the hypothesis on the level sets. I came up with the function $f(x)=(x^{-1/10}-1)\cdot\mathbf1_{x>1}(x)$ by reverse engineering the bound on the level sets. This function will not be in $L^{p}(\mathbb R)$ for all $p>9.$
Is my work above accurate? Any comments are welcomed, be it about style, lack of details, and of course, correctness of the ideas.
Thank you for your time.