My Definitions. Let $G$ be a locally compact Hausdorff group and $\mu$ be a Haar measure on $G$. We have defined "Haar" measure as follows:
(Haar measure) It's a nonzero left invariant outer Radon measure on $G$. An outer Radon measure is a locally finite Borel measure on $G$ which is outer regular on Borel sets and inner regular on open sets
(Assuming this definition) the author of the text have used the following statement without proof. But I'm not very sure how exactly to prove this:
Statement 1. Let $f \in L^1(G)$ such that $$ \int_Gg(x)f(x) ~ \mathrm{d}\mu(x)=0 ~ \text{for all }g\in L^1(G) $$ Then $f=0$.
I tried to proceed with the same line of argument as given in my earlier post this: since each compact set has finite measure, the assumption on $f$ implies that $$ \int_K f(x)~\mathrm{d}\mu(x)=0 ~\text{for all compact }K\subset G ~~~~~~~ (*) $$
Case 1. $f\in C_c(G)$ and $f$ is $\Bbb{R}$-valued.
In this case I have managed to prove $f=0$.For this let $S:=\{f\ne 0\}=\{f>0\}\cup\{f<0\}\subset G$. With contrary if $\mu(S)>0$ then WLOG $\mu(\{f>0\})=\cup_n \{f>\frac{1}{n}\}=:\cup_n E_n>0$ hence there must exists $N\ge 1$ s.t. $\mu(E_N)>0$. Since $f$ is continuous so $E_N$ is open in $G$. Now I'll use the (weak) inner regularity of $\mu$ to arrive at a contradiction: if $K$ be a compact subset of $E_N$ with $\mu(K)>0$ then we get $$ 0=\int_K f(x) ~\mathrm{d}\mu(x)\ge \int_K \frac{1}{N} ~\mathrm{d}\mu(x)=\frac{1}{N}\mu(K)>0 $$ a contradiction, hence $f=0$ in this case.
Case 2. $f\in C_c(G)$
This follows from Case 2.
Case 3. $f\in L^1(G)$. Here I got stuck, I know $C_c(G)$ is dense in $L^1(G)$. So there would exists a sequence of members in $C_c(G)$ converging towards $f$. But the members of that sequence need not satisfy $(*)$. Also even such sequence exists I don't know how to interchange the "limit" with "integral"??
Is the original statement valid only for continuous $f$? Thank you.
EDIT: If we look carefully the answer given by @Kavi Rama Murthy below, we can get a slightly general statement for "outer Radon measures" on a group $G$ as above, infact any $f\in L^1(G)$ satisfying the slight "general" condition $(*)$ above can be shown to be zero a.e. (provided $\mu$ is an outer Radon measure on $G$): Precisely,
Statement 2. Let $X$ is a locally compact Hausdorff space and $\mu$ is an outer Radon measure on $X$. Suppose $f\in L^1(\mu)$ s.t. $$ \int_K f(x)~\mathrm{d}\mu(x)=0 ~\text{for all compact }K\subset X $$ Then $f=0$ a.e.
Proof. As I've mentioned above the proof will be same as given in the answer below once we recall the following fact (which can be proved separately):
For an outer Radon measure $\mu$ on $X$ and every measurable set $A\subset X$ with $\mu(A)<\infty$ one has $$ \mu(A)=\sup \{\mu(K):K\subset A, K~\text{compact} \} $$
Assume this. Suppose WLOG $f$ is $\Bbb{R}$-valued. Then $\{f>0\}=\cup_n \{f>\frac{1}{n}\}=\cup_n A_n$. By Chebychev's inequality $\mu(A_n)<\infty$ for all $n$. Now if $K\subset A_n$ be a compact set with $\mu(K)>0$ then $$ 0=\int_K f \ge \frac{1}{n} \mu(K)>0 $$ a contradiction. Hence $\mu(K)=0$ for all compact $K$ with $K\subset A$. This implies $\mu(A_n)=\sup_{\text{such }K}\mu(K)=0$ and this is true for all $n$. Therefore $\mu(\{f>0\})=0$. $\square$
You are making things too complicated. You don't require any special property of Haar measure for this.
Let $A_n=\{g\in G: f(g)>\frac 1n\}$. Then $\mu (A_n) \leq n\int_G |f|d\mu$ so $\mu (A_n) <\infty$. Taking $g=\chi_{A_n}$ we see that $\int_{A_n} fd\mu=0$ which implies that $\mu (A_n)=0$. This is true for each $n$ so $\mu \{g: f(g) >0\}=0$. Similarly, $\mu \{g: f(g) <0\}=0$ so $f=0$ a.e.