Proving that $ f = 0 $ $\mu$-a.e. given $\int_O f \, d\mu = 0$ for all open sets $ O $

Question

I have the following problem:

Let $X$ be a topological space and $ \mu $ a Borel measure. Show that if $\int_O f \, d\mu = 0 $ for all open sets $ O$, then $f = 0$ $\mu$-almost everywhere.

I know the following related result:

Let $ f : [a, b] \to \mathbb{R} $ be integrable such that $ \int_a^x f(t) \, dt = 0 $ for all $ x \in [a, b] $. Then $ f(t) = 0 $ for almost every $ t $.

Proof:

It follows that $ \int_c^d f(t) \, dt = 0 $ for all $ a \leq c \leq d \leq b $. Hence, $ \int_O f \, dm = 0 $ for all open sets $ O $.

But I need to conclude that $ f = 0 $ $\mu$-almost everywhere. How can I make this final step in the proof?

Thank you for your help!

Answer 1

Consider $\{A\in \mathcal B(X): \int_A f d\mu=0\}$ (where $\mathcal B(X)$ is the Borel $\sigma-$algebra of $X$).

Verify that this is a $\lambda-$system. It contains all open sets. By the $\pi -\lambda$ theorem it follows that it contans all Borel sets. A standard argument now shows that $f=0$.a.e.

Ref. for $\pi -\lambda$ theorem: https://en.wikipedia.org/wiki/Dynkin_system

Answer 2

For an arbitrary Borel measure in a topological space, you have no option other than an abstract set theoratic proof. The reason being that you cannot use topological properties like that in the case of Real numbers. In addition, you miss out on the essential "Outer Regularity" which the Lebesgue Measure has which allows you to approximate measurable sets by open sets.

In the case of $\Bbb{R^{n}}$, for any given Borel measurable set $A$, you can find a sequence of open sets $O_{n}$ such that $A\subseteq O_{n}$ and $\lambda(O_{n}\setminus A)\to 0$

So let $G=\bigcap_{n=1}^{\infty}O_{n}$ . Then $A\subseteq G$ and $\lambda(G\setminus A)=0$

Now note that $\bigcap_{k=1}^{n}O_{k}$ is an open set and $\mathbf{1}_{\bigcap_{k=1}^{n}O_{k}\setminus A}\to \mathbf{1}_{G\setminus A}$ and $\mathbf{1}_{G\setminus A}=0$ almost everywhere.

Also as $\bigcap_{k=1}^{n}O_k$ is open, you have $\int_{\bigcap_{k=1}^{n}O_{k}}f\,d\lambda=0$

So you have that $|\int_{A}f\,d\lambda|=|\int_{A}f\,d\lambda-\int_{\bigcap_{k=1}^{n}O_{k}}f\,d\lambda|$

and $$|\int_{\bigcap_{k=1}^{n}O_{k}\setminus A}f\,d\lambda| =|\int_{\Bbb{R^{n}}}f\cdot\mathbf{1}_{\bigcap_{k=1}^{n}O_{k}\setminus A}\,d\lambda|$$

But you have $f\cdot\mathbf{1}_{\bigcap_{k=1}^{n}O_{k}\setminus A}\to 0$ almost everywhere and $|f\cdot\mathbf{1}_{\bigcap_{k=1}^{n}O_{k}\setminus A}|\leq |f|$ such that $\int_{\Bbb{R}^{n}}|f|\,d\lambda<\infty$.

Dominated Convergence Theorem gives you that $\int_{A}f\,d\lambda=0$.

Now you take $A=\{f>\frac{1}{n}\}$ to get that $\lambda(A)=0$. This holds for all $n$ and hence $f\leq 0$ almost everywhere

Similarly, you take $A=\{f<\frac{1}{n}\}$ and you'll get $f\geq 0$ almost everywhere.

For the general case, there's no easy way out other than the Dynkin's $\pi-\lambda$ Theorem or the Monotone Class Theorem as geetha290krm's answer does

Essentially you are showing that the collection $\mathcal{A}=\{A:\int_{A}f\,d\mu=0\}$ is equal to the Borel sigma algebra. You do that by showing it's a $\lambda$ system. Then by using the $\pi-\lambda$ theorem, you also have that $\mathcal{A}$ is a sigma algebra. But since $\mathcal{A}$ already contains open sets, it must contain the Borel sigma algebra (as Borel sigma algebra is the smallest sigma algebra containing the open sets). Thus $\mathcal{A}$ equals the Borel sigma algebra.

Now you again let $A=\{f>\frac{1}{n}\}$ and conclude as before.