Suppose $g$ satisfies equation $G(f)= \displaystyle{\int fg d \mu}$ for all $f$ in $L_1$ and that $c>1$. Let $E_c=\{ x: |g(x)|\geq c||G|| \}$, and define $f_c(x)$ to be $\pm1$ when $\pm|g(x)|\geq c||G||$ and to be $0$ when $x \notin E_c$. Then
$c||G|| \mu(E_c) \leq G(f_c) \leq ||G|| \mu(E_c) $,
which is a contradiction unless $\mu (E_c)=0$. Infer that $|g(x)| \leq ||G||$ for $\mu$-almost all $x$.
I have no idea how to solve this exercise. Can anyone help? The definition of $||G||$ is $||G||=Sup \{ |G(f)|: f \in L_p, ||f||_p = 1 \}$ and $G$ is a linear functional. This is the exercise 8.T of the book Bartle, The Elements of Integration and Lebesgue Measure.
The case $||G||=0$ obviously implies that $g=0\ a.e.$ and all the corresponding results trivially hold. So here we assume $||G||\not=0$.
$$G(f_c)=\int f_cg\ d\mu=\int g\chi_{\{x: g(x)≥c||G||\}}\ d\mu\ +\ \int -g\chi_{\{x: -g(x)≥c||G||\}} \ d\mu$$ Both integrand on RHS are non-negative. So we can apply Chebyshev's inequality. $$\int g\chi_{\{x: g(x)≥c||G||\}}\ d\mu\ ≥ c||G||\ \mu(\{x: g(x)≥c||G||\})$$ and $$\int -g\chi_{\{x: -g(x)≥c||G||\}}\ d\mu\ ≥ c||G||\ \mu(\{x: -g(x)≥c||G||\})$$ Therefore $$G(f_c)≥c||G||\mu(E_c)$$
Next note that since $G$ is a linear functional we have $$|G(f_c)|≤||G||\ ||f_c||=||G||\int |f_c|\ d\mu=||G||\ \mu(E_c)$$
Hence we get $$c||G||\mu(E_c)≤G(f_c)≤||G||\mu(E_c)$$ and this implies $c||G||\mu(E_c)≤||G||\mu(E_c)$ since $c>1,||G||\not=0$ measure of $E_c$ must be $0$. Therefore $|g(x)|≤||G||$ a.e. $(\mu)$.