# Show that $|v|(A) = \int_A |g(t)| dt$ for all Borel subsets $A$ of $[0,1].$

I am attempting the following question from 'An Introductory Course in Functional Analysis' by Bowers and Kalton, Chapter $2,$ Exercise $2.4.$

Question: Let $g \in L_1(0,1),$ the Banach space of $L_1$- functions on $[0,1]$ with Lebesgue measure, and define a measure on $[0,1]$ by $$v(A) = \int_A g(t) dt,$$ where $A$ is a Borel subset of $[0,1].$ Show that $$|v|(A) = \int_A |g(t)| dt$$ for all Borel subsets $A$ of $[0,1].$

The total variation $|v|$ of measure $v$ is defined as $$|v|(A) = \sup\{ \sum_{j \in F} |v(A_j)|:(A_j)_{j \in F} \text{ is a finite measurable partition of }A \} .$$

My attempt:

For each finite measurable partition $(A_j)_{j \in F}$ of $A$, we have $$\sum_{j \in F}|v(A_j)| = \sum_{j \in F}\left|\int _{A_j}g(t)dt \right| \leq \sum_{j \in F}\int_{A_j} |g(t)| dt.$$ That's all I can obtain.

Any hint would be appreciated.

• If $g$ is real-valued, let $A_1 = \{ x \in A : g(x) \geqslant 0\}$ (choose a Borel-measurable representative of the equivalence class of $g$). If $g$ is complex valued, it gets a bit more complicated and you'll need a sequence of partitions with increasingly many members. – Daniel Fischer Jun 24 '17 at 13:00
• @DanielFischer: Can you elaborate more on your answer? Why do we need to consider $A_1?$ – Idonknow Jun 25 '17 at 16:07
• Because $\nu(A_1) = \int_{A_1} g(t)\,dt = \int_{A_1} \lvert g(t)\rvert\,dt$. What is $\nu(A_2)$ where $A_2 = A \setminus A_1$? – Daniel Fischer Jun 25 '17 at 18:22
• @DanielFischer: I have posted an answer, can you check whether it is correct? Thanks. – Idonknow Jun 26 '17 at 4:15

Since $$\sum_{j \in F}|v(A_j)| = \sum_{j \in F}\left|\int _{A_j}g(t)dt \right| \leq \sum_{j \in F}\int_{A_j} |g(t)| dt = \int_A|g(t)|dt$$ for any partition $\{A_j\}$ of $A,$ we have $$|v|(A) \leq \int_A |g(t)| dt.$$
Let $A_1 =\{ x\in A: g(x) \geq 0 \}$ and $A_2 = \{ x \in A: g(x) < 0 \}.$
Observe that $\{ A_1,A_2 \}$ is a partition of $A$ and $$|v(A_1)| + |v(A_2)| = \left|\int_{A_1} g(t) dt \right| +\left| \int_{A_2} g(t)dt \right| = \int_{A_1} g(t) dt - \int_{A_2} g(t) dt = \int_{A_1} |g(t)| dt + \int_{A_2} |g(t)| dt.$$
Hence, $|v|(A) = \int_A |g(t)|dt.$