Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue measurable function and $\phi(x)=\lambda ( \lbrace{ t: f(t) >x \rbrace} )$. Prove that $\phi$ is right-continuous but not necessarily left-continuous.
I was thinking this: to every measure $\mu$ I can associate a right-continuous monotone function such that if $\mu([a,b]) < \infty$ then $\mu([a,b])=F(b)-F(a)$. In this case, the Lebsesgue measure, the function would be the identity function. Is this argument related to the exercise? And if yes how can I continue?
Let $x\in\mathbb{R}$ be fixed, and consider a sequence $(x_n)_{n\in\mathbb{N}}$ that tends to $x$ by above.
Then,
$$\left|\phi(x)-\phi(x_n)\right|=\lambda\left(\left\{t\ :\ x<f(t)\le x_n\right\}\right)=\lambda\left(f^{-1}(]x,x_n])\right).$$
Now, the sequence $(]x,x_n])_{n\in\mathbb{N}}$ is a decreasing sequence of Borel intervals, hence $(f^{-1}(]x,x_n]))_{n\in\mathbb{N}}$ is a decreasing sequence of Lebesgue-measurable sets.
Note that $\bigcap_{n\in\mathbb{N}}f^{-1}\left(]x,x_n]\right)=f^{-1}\left(\bigcap_{n\in\mathbb{N}}]x,x_n]\right)=\emptyset$, hence by continuity of $\lambda$ at $0$,
$$\lim_{n\rightarrow+\infty}\left|\phi(x)-\phi(x_n)\right|=\lim_{n\rightarrow+\infty}\lambda\left(f^{-1}(]x,x_n])\right)=\lambda\left(\bigcap_{n\in\mathbb{N}}f^{-1}(]x,x_n])\right)=0.$$
To prove that the converse does not necessarily hold, consider $f=\widetilde{x}\in\mathbb{R}$ which is constant.
Let us show that $\phi$ is not left-continuous at $\widetilde{x}$. Consider a sequence $(x_n)_{n\in\mathbb{N}}$ that tends to $\widetilde{x}$ by below.
Similarly, for any $n\in\mathbb{N}$,
$$\left|\phi(\widetilde{x})-\phi(x_n)\right|=\lambda\left(\left\{t\ :\ x_n<f(t)\le \widetilde{x}\right\}\right)=\lambda\left(\mathbb{R}\right)=+\infty,$$
which can thus not tend to zero.