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.


1 Answer 1


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 $fg = |f|$.

  • $\begingroup$ Many thanks! That was so clever! $\endgroup$
    – Suiz96
    Feb 2, 2019 at 8:44

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