$L^1$ is weak-star dense in the space of regular Borel measures

Question

Recall that you can identify the space $L^1[0,1]$ via the map $$ L^1[0,1] \to M[0,1], \quad f \mapsto \mu, $$ where $$ \mu(A) := \int_A f(x) \, \mathrm d x \quad \text{for each Borel measurable set } \ A \subseteq [0,1]$$ with a closed subspace of the Banach space $M[0,1]$ of regular Borel measures. Moreover, $M[0,1]$ is the dual space of $C[0,1]$ and therefore we can equip $M[0,1]$ with the weak*-topology from this duality.

Now I recently stumbled upon the following statement: $L^1[0,1]$ is dense in $M[0,1]$ with respect to the weak*-topology.

I have a clue how to prove it but I wonder if it is the usual way to do it: I know that you can approximate Dirac measures in the weak*-topology by functions from $L^1[0,1]$: Just choose a Dirac sequence $(\varphi_n)_{n \in \mathbb N}$ in $L^1[0,1]$ and observe that $$\int_0^1 f \varphi_n \longrightarrow f(0) = \delta_0(f)\quad \text{as } \ n \to \infty$$ for each $f \in C[0,1]$. By shifting the argument of each $\varphi_n$ appropiately one can approximate also every other Dirac functional in the weak*-star topology. But the set of Dirac functionals in dense in the weak*-topology as a result of the Hahn-Banach theorem. Hence $L^1[0,1]$ is weak*-dense in $M[0,1]$.

I am aware that this question is kind of subjective and ambigious, but is there a "more direct" way to prove it? My proof also does not offer any quantative information about the norms of the approximated measure! I would be also glad for a citable reference if there is one.

Answer 1

Let $\mu$ be a finite Borel measure on $[0,1]$ For each $n$ and $1 \le i \le n$, let $A_{i,n}$ be the half-open interval $[(i-1)/n, i/n)$, with $A_{n,n} = [(n-1)/n, 1]$ as a special case. Consider the step function $$g_n = \sum_{i=1}^n n \mu(A_{i,n}) 1_{A_{i,n}}.$$ Now if $f \in C([0,1])$, we have $$\int_0^1 f(x) g_n(x)\,dx = \sum_{i=1}^{n} n \mu(A_{i,n}) \int_{(i-1)/n}^{i/n} f(x)\,dx = \int_0^1 \sum_{i=1}^{n} \left( n \int_{(i-1)/n}^{i/n} f(x)\,dx\right) 1_{A_{i,n}}\,d\mu. $$ Now show that $h_n := \sum_{i=1}^{n} \left( n \int_{(i-1)/n}^{i/n} f(x)\,dx\right) 1_{A_{i,n}}$ converges pointwise to $f$ (use the uniform continuity of $f$) and is bounded by $\sup |f|$. Hence by dominated convergence we have $\int_0^1 f(x) g_n(x)\,dx = \int h_n\,d\mu \to \int f\,d\mu$ which shows the desired weak-* convergence.

In a certain sense, what we are doing is restricting $\mu$ to the smaller $\sigma$-algebra generated by the partition $\{A_{1,n}, \dots, A_{n,n}\}$. On this $\sigma$-algebra, it is absolutely continuous to Lebesgue measure, and $g_n$ is the Radon-Nikodym derivative. As the $\sigma$-algebras get finer and finer, they "approach" the Borel $\sigma$-algebra.

You could adjust this construction a little bit and show, for instance, that $C^\infty([0,1])$ is also weak-* dense in the Borel measures.

Answer 2

This follows easily from some basic results from FA. The dual of the locally convex space $M[0,1]$ with weak$^{*}$ topology is nothing but $C[0,1]$. So the statement that $L^{1}$ is dense in this space is equivalent to the fact that if $g \in C[0,1]$ vanishes on $L^{1} [0,1]$ then it vanishes on $M[0,1]$. But $g$ continuous and $\int f(x)g(x)=0$ for all $f \in L^{1} [0,1]$ implies $g(x)=0$ for all $x$ so $\int g d\mu=0$ for all $\mu$.