If $f\in L^1(\mathbb{R})$, I would like to prove or disprove that $$\lim_{\lambda\to\infty}\int_{|f|\ge \lambda}|f|\,dx=0.$$ I believe it is true, though I am not certain, as I have verified it for various integrable functions, such as the set $\{x^{-\alpha}\chi_{[0,1]}: 0<\alpha<1\}\subset L^1(\mathbb{R})$. Moreover, it fails to be true for $x^{-1}\chi_{[0,1]}$, which is indeed not integrable.
The result is of course true if $|f|$ is bounded, and I know that I can approximate arbitrarily well $f$ in $L^1$ by test functions $\psi\in C_c^\infty$ (given $\varepsilon>0$, choose $\psi$ such that $|\psi-f|_{L^1)<\varepsilon$), but I then am left with the following $$\int_{|f|\ge \lambda}|f|\,dx< \varepsilon + \int_{|f|\ge \lambda} |\psi|\,dx$$ and I am not sure how the handle the rightmost term.
It's true (in any measure space $(X,\Sigma;\mu))$. Maybe the easiest way to see this is to consider the sequence $|f|\cdot\chi_{|f|>\lambda_n}$ (where $(\lambda_n)_{n\in\Bbb N}$ is a sequence of positive reals that diverges to $+\infty$) and note this sequence is dominated by $|f|\in L^1$, so the D.C.T (or an easy adapation of the M.C.T) applies to say: $$\lim_{n\to\infty}\int_{|f|>\lambda_n}|f|\,\mathrm{d}\mu=\int_{f^{-1}\{\pm\infty\}}|f|\,\mathrm{d}\mu=0$$
Noting that if $f$ is integrable, $\mu(f^{-1}\{\pm\infty\})=0$ is forced.
An "alternative" method: integration by $f$ is absolutely continuous. From $\mu(f^{-1}\{\pm\infty\})=0$ and $\mu(|f|^{-1}(\lambda_n,\infty))<\infty$ for all $n$ (else, $\int_X|f|\,\mathrm{d}\mu\ge\lambda_n\cdot\mu(|f|^{-1}(\lambda_n,\infty))=\infty$ which contradicts integrability) we can use continuity of measure to infer that $\mu(|f|^{-1}(\lambda_n,\infty))$ can be made arbitrarily small for large $n$. Coupling with absolute continuity of the integral, we can deduce the integral of $|f|$ over the arbitrarily small measure sets $\{|f|>\lambda_n\}$ can also be made arbitrarily small.
This is not really an alternative since iirc absolute continuity of the integral is proven using either the M.C.T or D.C.T, but it's nice.