Claim $\limsup_{k\rightarrow \infty}f_k\leq f$ in "Measure theory and fine properties of functions" In the book "Measure theory and fine properties of functions", the author is proving that $D_{\mu}\nu$ is measurable when $\mu$ and $\nu$ are Radon measures. He claims that $\limsup_{y\rightarrow x}\mu(B(y, r))\leq \mu(B(x, r))$ where I guess $B(x, r)$ is closed ball. He takes $f_k=\chi_{B(y_k, x)}$, where $y_k\rightarrow x$ and $f=\chi_{B(x,r)}$. Then he just says $$\limsup_{k\rightarrow \infty}f_k\leq f.$$
I don't understand this step, would someone kindly explain me this.
Also, if the ball $B(x, r)$ is open, then why $$\liminf_{k\rightarrow\infty}f_k\geq f$$ holds?
Thank you in advance.
 A: Let $d$ be the metric on the space. Distinguish between the cases $d(x,z) < r$, $d(x,z) = r$ and $d(x,z) > r$ to investigate the behaviour of $f_k(z)$.
If $d(x,z) < r$, then we also have $d(y_k,z) < r$ for all sufficiently large $k$, thus $\lim\limits_{k\to\infty} f_k(z) = 1 = f(z)$. Similarly, if $d(x,z) > r$, then $d(y_k,z) > r$ for all sufficiently large $k$, and $\lim\limits_{k \to \infty} f_k(z) = 0 = f(z)$. In fact, these two cases can be handled at once: Let $\varepsilon = \lvert r - d(x,z)\rvert$. When $k$ is so large that $d(x,y_k) < \varepsilon$, then we have $f_k(z) = f(z)$. Thus
$$d(x,z) \neq r \implies f(z) = \lim_{k \to \infty} f_k(z).$$
It remains to look at the sphere $\{ z : d(x,z) = r\}$. But trivially, since the $f_k$ are characteristic functions, we have
$$0 \leqslant \liminf_{k\to \infty} f_k(z) \leqslant \limsup_{k \to \infty} f_k(z) \leqslant 1$$
for every $z$. If $B(x,r)$ denotes the closed ball $\{ p : d(x,p) \leqslant r\}$, then we have $f(z) = 1$ on the sphere, and $\limsup\limits_{k \to \infty} f_k(z) \leqslant f(z)$ for every $z$. If $B(x,r)$ denotes the open ball $\{ p : d(x,p) < r\}$, then we have $f(z) = 0$ on the sphere, and $\liminf\limits_{k \to \infty} f_k(z) \geqslant f(z)$ for every $z$.
