One of the properties of Lebesgue integrable functions, as stated here problem 3.7, is $f\in \mathcal{L}^1$ iff $\int|f|d\mu<\infty$ ,where $\mathcal{L}^1$ is the family of all integrable functions.
I tried proving this by: $$f\in \mathcal{L}^1\implies \int f^+d\mu<\infty \text{ and } \int f^-d\mu<\infty$$ hence $$\int |f|d\mu=\int |f^+-f^-|d\mu$$
And here I don't know how can I break up the integral. on the other hand: $$\int |f|d\mu<\infty\implies\int |f|d\mu=\int |f|^+d\mu-\int |f|^-d\mu=\int |f|^+d\mu=\int f^+d\mu<\infty$$ Is this any good?
How do I show both sides?
By definition, $$f \in \mathcal L^1 \quad \text{ iff } \quad \int f^+ d\mu < \infty \text{ and } \int f^- d\mu < \infty.$$ Now assume $f \in \mathcal L^1$. Since $|f| = f^+ + f^-$, we get $$\int |f| d\mu = \int f^+ + f^- d\mu = \int f^+ d\mu + \int f^- d\mu$$ by linearity, and both of these are by definition $< \infty$, so $\int |f|d\mu < \infty$.
For the opposite direction, we will prove the contrapositive. Assume that $f \notin \mathcal L^1$. Again we get $$\int|f|d\mu = \int f^+ d\mu + \int f^- d\mu,$$ but since $f \notin \mathcal L^1$, one of two summands on the right side must be $\infty$. Hence $\int |f|d\mu = \infty$.