Must $f$ necessarily be integrable?

Question

Let $c \in \mathbb{R}_{>0}$. Let $f$ be a non-negative, real-valued, measurable function defined on $[0,1]$ satisfying $\lambda \cdot m(\{x \in [0,1]:f(x)\geq \lambda\})\leq c/ \lambda$ for each $\lambda >0$.

My question is, must $f$ necessarily be integrable?

I know that $ \lambda \cdot m(\{x \in [0,1]:f(x)\geq \lambda\}) \leq \int_Ef$ by Chebyshev's inequality. However, I don't think this helps because the inequality is the "wrong" way. Therefore, I suspect that $f$ is not required to be integrable, although I can't think of a counterexample. I tried setting $f(x)=\begin{cases} 1/x,x \neq0 \\ 0,x=0 \end{cases}$, which I already know is not integrable, but this function has $\lambda \cdot m(\{x \in [0,1]:f(x)\geq \lambda\})=1 \nleq c/ \lambda$ for each $\lambda$ sufficiently large.

Answer 1

Weak type $L^{2,\infty}$ is embedded into $L^{q}$ for $0<q<2$ in finite measure space: \begin{align*} \int_{E}|f(x)|^{q}d\mu(x)\leq\dfrac{2}{2-q}\mu(E)^{1-q/2}\|f\|_{L^{2,\infty}}^{q}, \end{align*} this is an exercise in Loukas Grafakos book Classical Fourier Analysis, page 14. Essentially, you need a delicate inequality: \begin{align*} \mu(E\cap(|f|>\alpha))\leq\min\left(\mu(E),\alpha^{-2}\|f\|_{L^{2,\infty}}^{2}\right). \end{align*}

Answer 2

$f$ is necessarily integrable. $\int_0^{1} f(x)dx=\int_0^{\infty} m( \{x: f(x) >\lambda\}) d\lambda \leq \int_0^{1} 1 d\lambda +\int_1^{\infty} \frac c {\lambda^{2}}d\lambda <\infty$.