Embedding L1 in the Banach space of complex Borel measures

[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.

 

[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.

 

[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.