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Let $X$ be a Hausdorff locally compact space and let $\mu$ and $\lambda$ be two regular complex Borel measures in $X$. I'm trying to proof the following questions:

  1. $\mu + \lambda$ is also a regular complex Borel measures in $X$.
  2. $|\mu + \lambda| \leq |\mu| + |\lambda|$.
  3. $L^1(|\mu|) \cap L^1(|\lambda|)$ is dense in $L^1(|\mu + \lambda|)$.

For a complex measure $\mu$ we define $|\mu| = \inf\left \{ \sum_{n=1}^\infty |\mu(E_n)|: (E_n) \text{ is a partion of measurables of } E \right \}$.

I have a problem in the questions 1 and 3. In the first question, It's not difficult to proof that $\mu + \lambda$ is a complex measure, however I'm stuck in the regular part.

In the third question, using 2 we have that $L^1(|\mu|) \cap L^1(|\lambda|) \subset L^1(|\mu + \lambda|)$. I need some hint to proof the density part.

Help?

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I think $\lambda$ is supposed to be positive finite measure here. Anyway, $L^{1} (\lambda)$ can only mean $L^{1} (|\lambda|)$ and regularity w.r.t. $\lambda$ is same as regularity w.r.t. $|\lambda|$ so assume that $\lambda$ is a positive measure. Hint for 3): any bounded measurable function is integrable w.r.t. a complex measure, so given $f \in L^{1} (|\lambda+\mu|)$ consider the sequence $fI_{\{x:|f(x)| \leq n\}}$.

Hint for 1): $|\mu|$ and $|\lambda|$ are regular so $|\mu|+|\lambda|$ is regular. Regularity of $\mu +\lambda$ is now obvious from definition since $|\mu +\lambda| \leq |\mu|+|\lambda|$.

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  • $\begingroup$ By a complex measure I mean a measure $\lambda: X \to \mathbb C$. I've correct in the question the $L^1(|\lambda|)$. $\endgroup$ – user 242964 Jun 5 at 23:28
  • $\begingroup$ Yes, that is the definition I have used in my answer. $\endgroup$ – Kavi Rama Murthy Jun 5 at 23:29
  • $\begingroup$ Using the hint you gave for the first part, I got: for all $E$ measurable and $\epsilon > 0$, exists $V$ an open set such that $E \subset V$ and $(|\mu| + |\lambda|) (E) + \epsilon \geq (|\mu| + |\lambda|)(V)$ which is greater than $|\mu + \lambda| (V)$. For this, how can you see that $|\mu + \lambda| (E) + \epsilon \geq |\mu + \lambda|(V) $? $\endgroup$ – user 242964 Jun 7 at 13:15
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    $\begingroup$ @user242964 Yes, $I_A$ stands for the characteristic function of $A$. $\endgroup$ – Kavi Rama Murthy Jun 7 at 23:10
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    $\begingroup$ @user242964 First part: $|\mu+\lambda| (V)-|\mu+\lambda| (E)=|\mu+\lambda| (V\setminus E)\leq (|\mu|+\lambda|) (V\setminus E) <\epsilon$. $\endgroup$ – Kavi Rama Murthy Jun 7 at 23:13
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This may seem like a nitpick, but I actually think it's important because mathematics is nothing without precision. $L^1(|\mu|)\cap L^1(|\nu|)$, as written, is an intersection of collections of equivalence classes of functions and probably doesn't mean what you want it to mean. Indeed, if $\mu=0$ and $\nu\neq 0$ then $L^1(|\mu|)$ will contain exactly one equivalence class, namely the one consisting of all measurable functions. On the other hand, $L^1(|\nu|)$ contains multiple equivalence classes; in particular it does not contain the equivalence class consisting of all measurable functions. Consequently $L^1(|\mu|)\cap L^1(|\nu|)=\emptyset$! Instead of $L^1(|\mu|)\cap L^1(|\nu|)$ you probably meant $\{[f]:f\in\mathcal{L}^1(|\mu|)\cap \mathcal{L}^1(|\nu|)\}$ where $[\text{ }]$ denotes the equivalence class under the equivalence relation on $\mathcal{L}^1(|\mu+\nu|)$ induced by $|\mu+\nu|$-a.e. agreement (that an $f\in\mathcal{L}^1(|\mu|)\cap \mathcal{L}^1(|\nu|)$ will also lie in $\mathcal{L}^1(|\mu+\nu|)$ follows from the fact that $|\mu+\nu|\leq |\mu|+|\nu|$).

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