# A question about regular complex Borel measures

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?

## 2 Answers

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 $$|\mu +\lambda| \leq |\mu|+|\lambda|$$.

• By a complex measure I mean a measure $\lambda: X \to \mathbb C$. I've correct in the question the $L^1(|\lambda|)$. – user 242964 Jun 5 at 23:28
• Yes, that is the definition I have used in my answer. – Kavi Rama Murthy Jun 5 at 23:29
• 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 $|\mu + \lambda| (E) + \epsilon \geq |\mu + \lambda|(V)$? – user 242964 Jun 7 at 13:15
• @user242964 Yes, $I_A$ stands for the characteristic function of $A$. – Kavi Rama Murthy Jun 7 at 23:10
• @user242964 First part: $|\mu+\lambda| (V)-|\mu+\lambda| (E)=|\mu+\lambda| (V\setminus E)\leq (|\mu|+\lambda|) (V\setminus E) <\epsilon$. – Kavi Rama Murthy Jun 7 at 23:13

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 $$|\mu+\nu|\leq |\mu|+|\nu|$$).