$f$ is lebesgue integrable iff $f_+$ and $f_{-}$ are Lebesgue integrable

Question

I am doing some problems from the book "Theorems and problems in Functional Analysis" by A.A. Kirillov and A.D. Gvishiani.

The problem is that

For any real function $f:[0,1]\to\mathbb{R}$ and $$f_+(x):=\frac{f(x)+|f(x)|}{2}\,\,\,\,\,\,f_{-}(x):=\frac{|f(x)|-f(x)}{2}$$

Then $f$ is Lebesgue integrable iff $f_+$ and $f_-$ are Lebesgue integrable.

My reasoning is so far

$(\Rightarrow)$ Assume $f$ is integrable then $|f|$ is integrable and so are $f_+$ and $f_-$

$(\Leftarrow)$ Assume $f_+$ and $f_-$ are Lebesgue integrable, then $$f=f_+-f_-\hspace{0.5in} \text{After some rearrangement}$$ Hence $f$ is integrable.

Is my reasoning right? Also How do I show if $f$ is Lebesgue integrable then $|f|$ is also integrable. Any help would be highly appreciated. Thanks in advance.

Answer 1

It might be good practice to show some steps. For example, suppose $f$ is integrable, by which we mean $\int |f| d \mu < \infty$. Then, $$|f_+| = \frac{|f + |f||}{|2|} \leq |f|,$$ and integrating both sides shows that $f_+$ is integrable.

You seem to have the right idea for the other direction, but again, maybe you should write out the steps. How do you show that $\int |f|d\mu < \infty$?

For your last question, just note that $||f|| = |f|$.