Let $G$ be locally compact abelian, $T_2$ and $f\in L^1(G)$. Define $\mu(A)=\int_A f(x)\ dx$. Prove $\lVert \mu\rVert=\lVert f\rVert_1$

Question

Here $\mu$ becomes a complex measure and $\lVert \mu\rVert =|\mu|(G)$ is the total variation norm of $\mu$.

We have to show $|\mu|(G)=\int\limits_G |f(x)|\ dx$

Let $\{A_n\}$ be a partition of $G$. Then $\sum\limits_n |\mu(A_n)|\le\sum\limits_n \int\limits_{A_n} |f(x)|\ dx$

$=\sum\limits_n \int\limits_G |f(x)|1_{A_n}(x)\ dx$

$=\int\limits_{G}\sum\limits_n |f(x)|1_{A_n}(x)$ (by DCT)

$=\lVert f\rVert_1$

So $|\mu|(G)\le \lVert f\rVert_1$

But I cannot prove the other direction i.e. $|\mu|(G)\ge \lVert f\rVert_1$

If $f$ is real valued, then we can just take the partion $\{f\ge0\}$ and $\{f<0\}$, and our result will follow.

But I cannot prove it for general case. Can anyone help me in this regard? Thanks for help in advance.