Question: Let $f,g \colon \mathbb{R} \to \mathbb{R}$ be such that $g \in L_{loc}^1(\mathbb{R})$ and $f$ is measurable. If $\int_\mathbb{R} |f|^p |g| \, dx < \infty, 0 < p < \infty$ show that $$\lim_{t \to \infty} t^p \int_{\{|f| > t\}} g\, dx = 0\ \text{and}\ \lim_{t \to 0^+} t^p \int_{\{|f| > t\}} g\, dx =0.$$
My attempt: Define $\mu(dx) = |g| \, dx$ then $$\int_{\{|f| > t\}} g\, dx = \mu(\{|f| > t\}) = \mu(E_t)$$ where $E_t :=\{x \in \mathbb{R} \colon |f(x)|> t\}$. Since \begin{align*} + \infty > \int_{\mathbb{R}} |f|^p \, d \mu &= p \int_0^{\infty} t^{p-1} \mu(E_t)\, dt\\ &= p \int_0^1 t^{p-1} \mu(E_t)\, dt + p \int_1^\infty t^{p-1} \mu(E_t)\, dt\\ &= p \int_{\sqcup_{k\geq 0} F_k} t^{p-1} \mu(E_t)\, dt + p \int_{\sqcup_{k\geq 0} B_k} t^{p-1} \mu(E_t)\, dt \end{align*} where for each $k \geq 0$, $$F_k :=\{2^{-(k+1)} \leq t < 2^{-k}\}, B_k :=\{ 2^k < t \leq 2^{k+1}\}$$ Using monotone convergence theorem, we have $$+ \infty > p \sum_{k\geq 0} \int_{F_k} t^{p-1} \mu(E_t)\, dt,\quad +\infty > p \sum_{k\geq 0}\int_{B_k} t^{p-1} \mu(E_t)\, dt$$ So, $$\lim_{k\to \infty}\int_{F_k} t^{p-1} \mu(E_t)\, dt = \lim_{k\to \infty} \int_{B_k} t^{p-1} \mu(E_t)\, dt =0.$$ However, I am stuck at this step since I am not sure how to relate these two expression to the limit given in the question. Any hints/solutions are appreciated also can I check if $\mu$ is well-defined measure? As it is known that if $g$ is integrable $\mu$ is well-defined.
Let $\mu_f:=|g|\cdot dx$. Since $g\in L^{\operatorname{loc}}_1$, $\mu_g$ is a regular Borel measure (possible infinite) on $\mathbb{R}$.
For $t\rightarrow\infty$, the limit is rather straight forward if one uses Markov-Chebyshev's inequality:
$$t^p\int_{\{|f|>t\}}|g(x)|\,dx=t^p\mu_g(|f|>t)\leq\int_{\{|f|>t\}}|f|^p\,d\mu_g=\int_{\{|f|>t\}}|f|^p|g|\,dx\xrightarrow{t\rightarrow\infty}0$$ by dominated convergence. The result (for $t\rightarrow\infty$) follows from $$\Big|t^p\int_{\{|f|>t\}}g(x)\,dx\Big|\leq t^p\int_{\{|f|>t\}}|g(x)|\,dx, \qquad t\geq0$$
For $t\rightarrow0$, the argument is a little more laborious. Since $\mu_g$ is a regular Borel measure, the collection of $\mu_g$-integrable simple functions is dense in $L_p(\mu)$ for all $0<p<\infty$. Suppose $\phi=a\mathbb{1}_E$ with $\mu_g(E)<\infty$ and $a\in\mathbb{C}\setminus\{0\}$. As $\{|\phi|>t\}=E$ for $0<t<|a|$, $$t^p\mu_g(|\phi|>t)\xrightarrow{t\rightarrow0+}0$$ For $\phi=\sum^m_{n=1}a_n\mathbb{1}_{E_n}$, where $\mu_g(E_n)<\infty$ for all $1\leq n\leq m$, $$t^p\mu_g(|\phi|>t)\leq\sum^m_{n=1}t^p\mu_g(|a_n|\mathbb{1}_{E_n}>t/m)\xrightarrow{t\rightarrow0+}0$$
Here we use the fact that for functions $\phi_1,\ldots,\phi_m$ and numbers $b_1,\ldots, b_m$ $$\{\phi_1+\ldots+\phi_m > b_1+\ldots+ b_m\}\subset\bigcup^m_{j=1}\{\phi_j>b_j\}$$
Suppose that $\phi$ is a simple function such that $\|f-\phi\|^p_{L_p(\mu_g)}<\varepsilon$.
$$t^p\mu_g(|f-\phi|>t)\leq\int_{\{|f-\phi|>t\}}|f-\phi|^p\,d\mu_g\leq\|f-\phi\|^p_{L_p(\mu_g)}<\varepsilon$$
Then, from $$t^p\mu_g(|f|>t)\leq t^p\mu_g\big(|f-\phi|>t/2\big)+ t^p\mu_g(|\phi|>t/2)$$ it follows that $$\limsup_{t\rightarrow0+} t^p\mu_g(|f|>t)\leq2^p \varepsilon$$ and the conclusion for $t\rightarrow0+$ follows.