Equivalent assertions regarding an almost everywhere zero function on Lebesgue measure space

Question

This is an exercise question for my Measure Theory course and I cannot seem to be making any progress with the problem. Here it is:

Let $f \in L'(\mathbb{R})$ be a given function. Prove that the following assertions are equivalent.

a) $f = 0$ almost everywhere on $\mathbb{R}$.

b) $\int_K f(x) \,d\mu(x)= 0$ for each compact subset $K$ of $\mathbb{R} $

c) $\int_O f(x) \,d\mu(x) = 0$ for each open subset $O$ of $\mathbb{R}$

d) $\int_\mathbb{R} f(x)\omega(x) \,d\mu(x)= 0 $ for each function $\omega \in C_c(\mathbb{R})$, the space of the continuous functions with compact support.

e) $\int_\mathbb{R} f(x)\omega(x) \,d\mu(x)= 0 $ for each function $\omega \in C_0(\mathbb{R})$, the space of the continuous functions that vanish at infinity.

I suppose I have to show that $a \implies b \implies c \implies a$ and $ c \implies d \implies e \implies c $. So far I could show that $b \implies c$ but I have no clue how to show d or e. Any help would be greatly appreciated.

Answer 1

It is related to the regularity of Lebesgue measure. I prove $(c)\Rightarrow(a)$. Let $A=\{x\mid f(x)>0\}$ and $B=\{x\mid f(x)<0\}$. Supppose that $(c)$ holds. We go to prove that $\int_{A}f=0$ and it will follow that $\mu(A)=0$. (Prove by contradiction. Suppose the contrary that $\mu(A)>0$. Define $A_{n}=\{x\mid f(x)\geq\frac{1}{n}\}$. Observe that $A=\cup_{n}A_{n}$, so there exists $n$ such that $\mu(A_{n})>0$. Now $0=\int_{A}f\geq\int_{A_{n}}f\geq\frac{1}{n}\mu(A_{n})>0$, which is a contradiction.) Choose a decreasing sequence of open set $(O_{n})$ such that $A\subseteq O_{n}$ for each $n$ and $\mu(O_{n}\setminus A)\rightarrow0$ (see Remark 1 for the construction). Let $G=\cap_{n}O_{n}$, then $A\subseteq G$ and $\mu(G\setminus A)=0$. In other word, $1_{G}=1_{A}$ $\mu$-a.e.. Observe that $1_{O_{n}}\rightarrow1_{G}$ pointwisely and hence $f1_{O_{n}}\rightarrow f1_{G}$ pointwisely. Moreover, $(f1_{O_{n}})$ is dominated by the integrable function $|f|$. Now \begin{eqnarray*} \int_{A}fd\mu & = & \int_{G}fd\mu\\ & = & \int f1_{G}d\mu\\ & = & \lim_{n}\int f1_{O_{n}}d\mu\\ & = & \lim_{n}\int_{O_{n}}fd\mu\\ & = & 0. \end{eqnarray*}

Similarly, we can show that $\int_B f=0$ and hence $\mu(B)=0$.

Remark 1: Let $\varepsilon>0$ be given. For each $k\in N$, let $A_{k}=A\cap[-k,k]$. Note that $\mu(A_{k})<\infty$. Recall the way that we define Lebesgue (outer) measure of $A_{k}$: $\mu(A_{k})=\inf\{\sum_{i=1}^{\infty}\mu(I_{i})\mid A_{k}\subseteq\cup_{i=1}^{\infty}I_{i}\mbox{ and }I_{i} \mbox{ is an open interval.}\}$. There exists an open set $U_{k}$ such that $A_{k}\subseteq U_{k}$ and $\mu(U_{k})-\mu(A_{k})<\varepsilon/2^{k}$. Let $U=\cup_{k}U_{k}$. Note that $A=\cup_{k}A_{k}$. Then $A\subseteq U$ and \begin{eqnarray*} & & \mu(U\setminus A)\\ & \leq & \mu(\cup_{k}(U_{k}\setminus A_{k}))\\ & \leq & \sum_{k=1}^{\infty}\mu(U_{k}\setminus A_{k})\\ & \leq & \varepsilon. \end{eqnarray*} This shows that there exists a sequence of open sets $(U_{n})$ such that $A\subseteq U_{n}$ and $\mu(U_{n}\setminus A)\rightarrow0$. Finally, let $O_{n}=\cap_{i=1}^{n}U_{i}$, then $(O_{n})$ is a decreasing sequence of open sets with $A\subseteq O_{n}$ and $\mu(O_{n}\setminus A)\rightarrow0$.

Answer 2

Then I show $(d)\Rightarrow(c)$. Suppose that $(d)$ holds. We assert that $\int_{I}f=0$ for any open interval $I$ with $\mu(I)<\infty$. Let $I=(a,b)$ with $a<b$. Let $N\in\mathbb{N}$ be sufficiently large. For each $n\geq N$, construct a piecewise linear function $\omega_{n}$ such that:

1. $\omega_{n}=0$ on $(-\infty,a]\cup[b,\infty)$,

2. $\omega_{n}=1$ on $[a+\frac{1}{n},b-\frac{1}{n}]$,

3. $\omega_{n}$ is linear on $[a,a+\frac{1}{n}]$,

4. $\omega_{n}$ is linear on $[b-\frac{1}{n},b]$.

Clearly $\omega_{n}$ is a continuous function with compact support. Moreover, $\omega_{n}\rightarrow1_{I}$ pointwisely and $|\omega_{n}|\leq1$. By dominated convergence theorem, we immediately have $\int_{I}f=\lim_{n}\int f\omega_{n}=0$.

Next, we show that $\int_{I}f=0$ for any open interval $I$. This is easy. For, if $\mu(I)=\infty$, then $I$ can only take the form: $I=\mathbb{R}$, $I=(a,\infty)$, $I=(-\infty,b)$. For example, if $I=(a,\infty)$, then $1_{(a,a+n)}\rightarrow1_{I}$ pointwisely. By dominated convergence theorem again and the previous result, we have $\int_{I}f=\lim_{n}\int_{(a,a+n)}f=0$.

Finally, let $O$ be an open set, then $O=\cup_{i}I_{i}$, where $\{I_{i}\mid i=1,2,\ldots\}$ is a countable (finite or countably infinite) family of pairwisely disjoint open intervals. Note that $1_{O}=\sum_{i}1_{I_{i}}$, so $\int_{O}f=\sum_{i}\int_{I_{i}}f=0$. (Here dominated convergence theorem is used implicitly).