Application of Lebesgue Dominated Convergence. Let $f \in L^1(\mathbb{R})$, show that $$ \int_{-x}^{x} f(t) dt \to \int_{- \infty}^{\infty} f(t) dt$$ as $x \to \infty$. 
$\textbf{My Attempt:}$ We can rewrite, $$ \int_{-x}^{x} f(t) dt = \int_{\mathbb{R}} f(t) \chi_{(-x,x)} (t) dt$$
Let $\{x_n\}$ be a sequence of real numbers such that $x_n \to \infty$ as $n \to infty$, then $$ \lim_{x \to \infty} \int_{-x}^x f(t) dt = \lim_{n \to \infty} \int_{\mathbb{R}} f(t) \chi_{ (-x_n, x_n)} (t) dt$$ Let $g_n(x) = f(t) \chi_{(-x_n, x_n)}$, then $|g_n(x)|=|f(t) \chi_{(-x_n,x_n)}| \le |f(t)| \in L^1$. Also, as $n \to \infty$, we have $g_n(x) \to f(t) \chi_{(-\infty, \infty)}$. Then by Lebesgue dominated convergence we have, 
$$\lim_{x \to \infty} \int_{- x}^{x} f(t) dt = \int_{\mathbb{R}} \lim_{n \to \infty} f(t) \chi_{(-x_n, x_n)} (t) dt = \int_{- \infty}^{\infty} f(t) dt$$
Are there any errors in this proof? 

Your proof is fine.
As an alternative, you can also (writing $f = f_+ - f_-$) assume that $f\ge 0$; and then apply the monotonic convergence theorem.
