Help proving some lebesgue Integral Identity

Here is another solution for you. As you have notice that $$\int_{\mathbb{R}} |f(x)|dm(x) = \int_{\mathbb{R}} \left( \int_{0}^{|f(x)|} 1dm(t)\right)dm(x)=\int_{\mathbb{R}} \left( \int_{0}^{\infty} \mathbb{1} _\left\{\left(x,t\right) \mid 0\leq t\leq |f(x)| \right\} dm(t)\right)dm(x) $$ If we could have used tonelli's theorem we could have change the order of integration and get $$\int_{0}^{\infty} \left(\int_{\mathbb{R}} \mathbb{1} _\left\{\left(x,t\right) \mid 0\leq t\leq |f(x)| \right\} dm(t)\right)dm(x) = \int_{0}^{\infty} m\left( \left\{ x\in \mathbb{R} \mid|f(x)| \geq t \right\} \right)dm(t)$$ which is the desired equality. Therefore, it suffice to justify the usage of tonelli's theorem. The Lebesgue measure is complete and $\sigma$-finite thus it suffices to prove that the function $\mathbb{1}_\left\{ x\in \mathbb{R} \mid |f(x)| \geq t \right\}$ is a measurable function in the product space or, equivalently that the set $\left\{ x\in \mathbb{R} \mid |f(x)| \geq t \right\}$ is measurable.
Let $\alpha \in \mathbb{R}$. From the assumption f is lebesgue measurable (since is integrable) which implies that $|f|$ is also lebesgue measurable therefore the set $E_{\alpha} = \left\{x \in \mathbb{R} \mid |f(x)| > \alpha \right\}$ is measurable in $\left( \mathbb{R} , \mathcal{L} \left( \mathbb{R} \right) \right)$. $\mathbb{R}$ is obviously measurable and therefore the set $E_{\alpha} \times \mathbb{R}$ is a measurable rectangle which is by definition measurable in the product space. Define $$G(x,t) = |f(x)|$$and notice that $G^{-1} \left( \left(\alpha ,\infty \right)\right) = E_{\alpha} \times \mathbb{R}$. Since $\alpha$ was arbitrary this proves that $G$ is measurable. In the same way one can prove that the function $H(x,t) = t$ is measurable in the product space and therefore $$R(x,t) = G\left(x,t\right) - H\left(x,t \right) = |f(x)| - t$$ is measurable as a function from the product space to $\mathbb{R}$. Therefore the set $$\left\{ x\in \mathbb{R} \mid |f(x)| > t\right\} = R^{-1} \left( \left[0, \infty \right]\right)$$ is measurable in the product space.