I want to show $\dfrac{1}{x}\in L^{1,\infty}([0,1])$ and it fails in $L^1([0,1])$.
Here is my working :
Clearly,calculate by improper Riemann integral we have $\int_{0}^{1}\dfrac{1}{x} dx=\infty,$ so $\dfrac{1}{x}\not\in\ L^1([0,1])$
Now for any $\alpha>0,$ one has $\left\{x\in[0,1]:\dfrac{1}{x}>\alpha\right\}\subseteq[0,1]$, then $\left|\left\{x\in[0,1]:\dfrac{1}{x}>\alpha\right\}\right|\leq1$.
For each $0<\alpha\leq1,$ $$\alpha|\{x\in[0,1]:\dfrac{1}{x}>\alpha\}|\leq\alpha\cdot1=\alpha\leq1$$
For each $\alpha>1$ $$\alpha\left|\{x\in[0,1]:\dfrac{1}{x}>\alpha\}\right|=\alpha\left|\left\{x\in[0,1]:x<\dfrac{1}{\alpha}\right\}\right|\le\alpha\cdot\dfrac{1}{\alpha}=1$$
So,$$\left\lVert f\right\rVert_{L^{1,\infty}([0,1])}=\sup\bigg\{\alpha|\{x\in[0,1]:\dfrac{1}{x}>\alpha\}|:\alpha>0\bigg\}=1<\infty$$
So, our conclusion follows.
I will greatly appreciate it if anyone can take the time to check my working for correctness. Thanks for considering my request.
It looks good for me. In general, we have $\|f\|_{L^{p,\infty}}\leq\|f\|_{L^{p}}$ for $p\in[1,\infty]$ by Chebychev's inequality.