# Total variation of a finite signed measure defined by an integral

I am reading about total variation of signed measures. There is a proposition that I cannot complete it's proof:

Proposition. Suppose that $$(X,\mathcal{A},\mu)$$ is a measure space, that $$f\in\mathcal{L}_1(X,\mathcal{A},\mu,\mathbb{R})$$ and that $$\nu$$ is the finite signed measure defined by $$\nu(A)=\int_{A}fd\mu.$$ Then $$|\nu|(A)=\int_{A}|f|d\mu$$ holds for each $$A\in\mathcal{A}.$$

The proof is the next:

Let $$A\in\mathcal{A}$$ and let $$\{A_{n}\}_{n\in\mathbb{N}}$$ a measurable partition of $$A.$$ Then we have $$\sum_{n}|\mu(A_n)|=\sum_{n}|\int_{A_n}fd\mu|\leq\sum_{n}\int_{A_n}|f|d\mu=\int_{A}|f|d\mu.$$ Taking supreme over such sums we have the inequality $$|\nu|(A)\leq\int_{A}|f|d\mu.$$

To prove the other inequality I'd like to prove that there is a measurable sequence of simple functions $$\{g_n\}$$ such that $$|g_n(x)|=1$$ and $$\displaystyle\lim_{n \to{+}\infty}{g_n(x)f(x)}=|f(x)|$$ for each $$x\in X.$$ If such sequence existed we would have easily $$|\int_{A}g_nf d\mu|\leq |\nu|(A)$$ and for dominated convergence theorem we would have the desired inequality.

Any kind of help is thanked in advanced.

You don't even need to take a sequence to achieve this. If $$f$$ is measurable then $$A = \{f < 0\}$$ is a measurable set. In particular $$g = -1_{A} + 1_{A^c}$$ is a simple function with $$|g| = 1$$ and $$fg = |f|$$.