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I am learning Measure theory via self study of Bartle "The elements of Integration and Lebesgue Measure". I was stumped by the reasoning in one of the decomposition proofs. The point is to show that a Bounded Linear Functional can be represented as the difference of two positive bounded linear functionals. The proof presented is as follows.

Define $G^+ = sup\{G(f) : g \in L_p : 0 \le g \le f\}$ for all $f \ge 0$.

The next step is to ST $G^+$ is a BLF (Bounded Linear Functional).

It is clear that $G^+(cf) = c G^+ (f)$ for $c \ge 0$, $f \ge 0$ (This part was not a problem). The next step attempts to prove that given

$f_j \ge 0, G^(f_1 + f_2) = G^+(f_1) + G^+(f_2)$

This is where I got confused and did not quite understand the line of reasoning. This is how the proof continues:

If $0 \le g_j \le f_j$ Then $G(g_1) + G(g_2) = G(g_1 + g_2) \le G^+(f_1 + f_2)$

Taking supremum over all $g_j \in L_p$ we claim that

$G^+(f_1) + G^+(f_2) \le G^+(f_1 + f_2)$

I would have thought that it would be te other way around, i.e.

$sup\{G(g_1) : g_1 \in L_p, 0 \le g_1 \le f_1 \}$ + $sup\{G(g_2) : g_2 \in L_p, 0 \le g_2 \le f_2 \}$ >= $sup\{G(g_1 + g_2) : g_1, g_2 \in L_p, 0 \le g_j \le f_j\}$

Which implies $G^+(f_1) + G^+(f_2) \ge G^+(f_1 + f_2)$

The book continues: Conversely if $0 \le h \le f_1 + f_2$ let $g_1 = sup(h - f_2,0)$ and $g_2 = inf(h, f_2)$.

It follows that $g_1 + g_2 = h$ and that $0 \le g_j \le f_j$. Threfore

$G(h) = G(g_1) + G(g_2) \le G^+(f_1) + G^+(f_2)$

Since this is true for all $h \in L_p$ we have $G^+(f_1 + f_2) = G^+(f_1) + G^+(f_2)$

My questions were:

  1. Why would $G^+$ be positive?

  2. How did the final conclusion $G^+(f_1 + f_2) = G^+(f_1) + G^+(f_2)$ for f_1, f_2 in L_p follow?

I would have thougt that the definition for $G^+$ should have been

$G^+ = sup\{G(f) : g \in L_p : 0 \le g \le f, G(g) \ge 0\}$

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    I would like to add a question related to this problem: in the next step he defines $$G^+:M(X,\mathbb{X}) \to \mathbb{R}; f \mapsto G^+(f^+) - G^+(f^-)$$ claiming that this is the "positive part" of $G$, but I wonder why this is still positive2017-04-23

1 Answers 1