# Embedding L1 in the Banach space of complex Borel measures

1. Jun 11, 2012

### Undecided Guy

Hey, I know this is commonly a homework question, but it came up in my own studies; so this isn't a homework question for me. I hope it's alright that I put it here.

I'm trying to show that if $f dx = d\lambda$ for some $f \in L^1(\mathbb{R}^d)$ and complex Borel measure $\lambda$ then $|f| dx = d|\lambda|$ (i.e., $f \to fdx$ is an isometry).

What I've done so far is constructed a $\psi \in L^1(\lambda)$ so that $\psi d\lambda = d|\lambda|$ with $|\psi| = 1$. So we then have that $f \psi dx = \psi d\lambda = d|\lambda|$, so it looks like what I need to show is that $f \psi = |f|$. I'm not really sure how to do that though.

2. Jun 14, 2012

### Undecided Guy

Oh dear, sorry, what I meant is that I have a $\psi \in L^1(|\lambda|)$ so that $d\lambda = \psi d|\lambda|$. Sorry for the mistake.

Anyway, what I've ended up doing so far is showing one of the two ineqaulities. Fix a partition $\{X_j\}_{j=1}^{\infty}$ of $\mathbb{R}^d$.

Then $\sum_{j=1}^{\infty} |\lambda(X_j)| = \sum_j |\int_{X_j} f(x) dx | \le \sum_j \int_{x_j} |f| dx = \int_{\mathbb{R}^d} |f| dx = ||f||_1$

with the second to last equality holding from the monotone convergence theorem. Supping on all such partitions gives $|\lambda| \le ||f||_1$ directly from the definition. Still at kind of a loss how to show the reverse.

3. Jun 20, 2012

### Undecided Guy

For completeness, I thought I'd go ahead and update this with the solution.

First I'll need the following fact:

If $\mu$ is some positive measure, $g \in L^1(\mu)$, and for every measurable set $E$, $0 \le \int_E g(x) d\mu(x)$, then $g \ge 0$ almost everywhere (wrt $\mu$).

Proof: Consider the set $F = \{ g(x) \notin [0, \infty) \}$. Write $F = R \cup C$ for $R = \{ g(x) < 0 \}$ and $C = \{g(x) \in \mathbb{C} - \mathbb{R} \}$. F has positive measure if and only if at least one of R or C does as well.

Suppose R has positive measure. Write $R = \bigcup_j R_j$ for $R_j = \{ g(x) < -\frac{1}{j} \}$. Since R has positive measure, there's at least one j so that $R_j$ does as well. Thus $\int_{R_j} g(x) d\mu \le \int_{R_j} -\frac{1}{j} d\mu = -\frac{1}{j} \mu(R_j) < 0$, which contradicts the hypothesis on g. So R must have 0 measure.

Write g = u + iv. So on C, v is never 0. So $0 \le \int_C g(x) d\mu = \int_C u d\mu + i \int_C v d\mu$. This is only possible if $\int_C v d\mu = 0$. By writing $C = P \cup N$ where $P = \{v > 0 \}$ and $N = \{ v < 0 \}$ and applying the same argument as above on $R$, we see that both P and N must have 0 measure, so C does as well.

Together this shows that F must have 0 measure. //

Onto the actual statement, we want to show that:

If $\mu$ is a positive measure, $f \in L^1(\mu)$ and $\lambda$ is defined by $d\lambda = f d\mu$, then $||\lambda|| = |f| d\mu$

Proof: Construct a function $\psi \in L^1(|\lambda|)$ with $|\psi| = 1$ and $\psi d\|lambda| = d\lambda$. Now $\psi \bar{\psi} = |\psi| = 1$, so

$d|\lambda| = \bar{\psi} \psi d|\lambda| = \bar{\psi} d\lambda = \bar{\psi} f d\mu$.

Thus given any measurable set E, we have that $0 \le |\lambda|(E) = \int_E \bar{\psi} f d\mu$.

Appealing to the fact above, this implies that $\bar{\psi} f \ge 0$ almost everywhere (so modifying on a set of measure 0 and changing nothing in the above argument, we may assume it's nonnegative everywhere).

Thus $\bar{\psi} f = |\bar{\psi} f| = |f|$, since $|\bar{\psi} | = 1$. So, $d|\lambda| = \bar{\psi} f d\mu = |f| d\mu$, which gives the result.