Let $X$ be a set, $F$ a $\sigma$-field of subsets of $X$, and $\mu$ a probability measure on $X$. Given a random variable $f:X\rightarrow\mathbb{R}$, define $$\chi_f(t)=\int_Xe^{itf}d\mu$$ Show that $\chi_f$ is continuous and $|\chi_f(t)|\leq 1$.
I'm not sure how to start proving either statement. The function $\chi_f$ seems quite intangible, as it is an integral over $X$ and using $d\mu$.
The second part follows from $|e^{itf}|=1$.
For the first one, write $\chi_f(s)-\chi_f(t)=\int_Xe^{itf}(1-e^{i(s-t)f})\mathrm d\mu$. Define $f_n:=f\mathbb 1_{\{|f|\lt n\}}$; using the inequality $|e^{ia}-e^{ib}|\leqslant |a-b|$ we get $$|\chi_{f_n}(s)-\chi_{f_n}(t)|\leqslant \int_{\{f\lt n\}}|1-e^{i(s-t)f}|\mathrm d\mu\leqslant |s-t|n,$$ hence $f_n$ is uniformly continuous.
Now, we prove that $\chi_{f_n}\to \chi_f$ uniformly. We have $e^{itf}-e^{itf_n}=0$ if $|f(\omega)|\lt n$, hence $$|\chi_{f_n}(t)-\chi_f(t)|=\int_{\{|f|\geqslant n}|e^{itf}-1|\mathrm d\mu\leqslant 2\mu\{|f|\geqslant n\}.$$