$f$ is continuous at $x$ if and only if $\omega_f (x) = 0$ 
Prove that $f\colon[a,b]\to\Bbb R$ is continuous at $x\in[a,b]$ if and only if $\omega_f(x) = 0$.
  Here,
  $$\omega_{f}(x):= \lim _{\delta \rightarrow 0+} \Omega_{f}(B(x,\delta)\cap [a,b]),$$
  where
  $$\Omega_{f} (T) = \sup \{f(x) - f(y): x,y \in T\}$$
  is the oscillation of $f$ on $T \subset [a,b]$. 

I don't really understand the definition of $\omega_{f} (x)$. Please explain it to me and help me prove the statement.
Thanks.

In this case, $\omega_f(T)$ gives the degree to which $f$ 'oscillates' on the subset $T$ of its domain; it's equivalent to the difference between the supremum $\sup_{x\in T} f$ and $\inf_{x\in T} f$ on $T$. For interior points $x\in\operatorname{int}(T)$, the oscillation at a point $x\in T$ is defined as the limit of the oscillation of $f$ on progressively smaller balls $B(x,\delta)$ for sufficiently small $\delta>0$. The purpose of the intersection $B(x,\delta)\cap T$ is to only consider points within $T$ and is only significant here for determining the oscillation of $f$ at the boundary points $\partial T$.
Now, to show that a function $f$ is continuous at $x\in \operatorname{int}(T)$ iff the oscillation $\omega_f$ vanishes at $x$, consider what it means for $f$ to be continuous at $x$. Rigorously, we define $f$ to be continuous at $x$ iff for all $\varepsilon>0$, we can guarantee $f(y)\in B(f(x),\varepsilon)$ by requiring $y\in B(x,\delta)\subset T$ for some sufficiently small $\delta>0$. It follows that if $f$ is continuous then for all $\varepsilon$ we have that $|f(x)-f(y)|<\varepsilon$ for $y\in B(x,\delta)$ so thus $\omega_f(B(x,\delta))<\varepsilon$. Since we can constrain $\varepsilon$ to be arbitrarily small, it follows that we can constrain $\omega_f(B(x,\delta))$ to be as close to $0$ as we'd like for sufficiently small $\delta$ and thus $\omega_f(x)=0$.
Working in the reverse direction, suppose $\omega_f(x)=0$. Thus we have that we can make $\omega_f(B(x,\delta))=\sup_{y\in B(x,\delta)}\{f(x)-f(y)\}$ arbitrarily small by choosing sufficiently small $\delta$. Since $|f(x)-f(y)|\le\sup_{y\in B(x,\delta)}\{f(x)-f(y)\}$ for any $y\in B(x,\delta)$ this is equivalent to $f$ being continuous at $x$.
