Definition of weak* convergence in $L^{\infty}$ I'm new to functional analysis and have (what I think is) a basic question on the definition of weak* convergence.
Let $X$ be a normed linear space with dual $X^{\ast}$. According to the definition I know, a sequence $(f_n)_{n \in \mathbb{N}}$ in  $X^{\ast}$ is said to converge weakly* to $f \in X^{\ast}$ if $f_n$ converges pointwise to $f$, that is, if
$$\forall x \in X, \quad\lim_{n \rightarrow \infty} f_n(x) = f(x).$$
In the special case that $X^{\ast} = L^{\infty}(\Omega,\mu)$ for some measure space $(\Omega,\mathcal{A},\mu)$, I've stumpled upon a different definition (for example, here and here): a sequence $(f_n)_{n \in \mathbb{N}}$ in $L^{\infty}(\Omega,\mu)$ converges weakly* to $f \in L^{\infty}(\Omega,\mu)$ if
$$\forall g \in L^1(\Omega,\mu), \quad \lim_{n \rightarrow \infty} \int_{\Omega} f_n g \,\text{d}\mu = \int_{\Omega} f g \,\text{d}\mu.$$
My feeling is that both definitions are equivalent because of Riesz' Representation Theorem. Is this correct? If so, how can the equivalence be formally proved?

It's not the Riesz Representation Theorem, but one can show without a lot of effort that for sigma-finite measures the dual of $L^1(\Omega)$ can be identified with $L^\infty(\Omega)$ with the natural duality that makes $f\in L^\infty$ a linear functional on $L^1$ via
$$
g\longmapsto \int_\Omega fg.
$$
