I am new to measure theory and I have this problem: given a measurable function $f\colon J\subseteq R\rightarrow R$, show that the function $D_{f}(\lambda) = \mu(t\in J\colon\; f(t)\leq y)$ is monotone increasing and right-continuous. My question is about the proof that it is right-continuous. I have tried this:
$\lim_{u\to u_{0}^{+}} |D_{f}(u) - D_{f}(u_{0})| = \lim_{u\to u_{0}^{+}} \mu (t\in J \colon \; u_{0} < f(t)\leq u) = 0$.
However, the last equality seems true but I do not see a formal justification yet.
Let $(y_n) \searrow y$; we will show $\lim_{n \rightarrow \infty} D_f(y_n) = D_f(y)$. Note that $\{t : f(t) \le y_n\}$ is a decreasing sequence of sets, so by the monotone convergence theorem for sets we have \begin{align*} \lim_{n \rightarrow \infty} D_f(y_n) &= \lim_{n \rightarrow \infty} \mu(\{t : f(t) \le y_n\}) \\ &= \mu\left(\bigcap_{n=1}^\infty \{t : f(t) \le y_n\}\right) \\ &= \mu(\{t : f(t) \le y\}) \\ &= D_f(y). \end{align*}
Since $(y_n)$ was an arbitrary sequence decreasing to $y$ and $\lim_{n \rightarrow \infty} D_f(y_n) = D_f(y)$, we conclude $D_f(y)$ is right-continuous.