Deformation theorem and Tristan Needham's proof of Cauchy's formula

In pg 428 of Visual Complex Analysis, a proof of Cauchy's formula is being done and I am having the step of evaluating :

$$ \oint \frac{f(z)}{z-a} \frac{dz}{2\pi i} = <f>_{C_r}$$

My issue is in the remark he mentions of identifying the integral on left as the average of function over circle

Now shrink the cirlce $C_r$ towards its centre a, as illustrated in [1a]. Even if f is merely continuous (rather than analytic), $f(C_r)$ will shrink to $f(a)$ as indicated in $[1b]$... contin

How is it possible that we can apply the deform the curve if the function is only continuous? To my understanding integrals are same under deformation theorem only if the function is analytic while it is being deformed into the new region. Is there some other theorem which is being applied?

Analyticity is used to say that the integral doesn't depend on $r$. If $f$ is not analytic but merely continuous, then the average can depend on $r$, but it still approaches $f(a)$ as $r\to0$. When $r$ is near $0$, all the values of $f$ on the circle $C_r$ are close to $f(a)$, and therefore so is their average.

You're misunderstanding the statement being made. What Needham is saying is that $f$ being analytic implies that for a "nice loop" $L$ and any radius $r>0$, we have \begin{align} \frac{1}{2\pi i}\int_L\frac{f(z)}{z-a}\,dz&=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-a}\,dz \tag{Cauchy's theorem}\\ &=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(a+re^{it})}{re^{it}}ire^{it}\,dt\\ &=\frac{1}{2\pi}\int_0^{2\pi}f(a+re^{it})\,dt\\ &=\text{average of $f$ over $C_r$} \end{align} Since this is true for all $r>0$, we can now take limits to deduce \begin{align} \lim_{r\to 0^+}\frac{1}{2\pi i}\int_L\frac{f(z)}{z-a}\,dz&= \lim_{r\to 0^+}\left(\text{average of $f$ on $C_r$}\right) \end{align} Clearly, the LHS doesn't even depend on $r$, so we have \begin{align} \frac{1}{2\pi i}\int_L\frac{f(z)}{z-a}\,dz&= \lim_{r\to 0^+}\left(\text{average of $f$ on $C_r$}\right) \end{align} Finally, he is saying that $f$ analytic implies in particular $f$ is continuous, and that continuity of $f$ (at $a$) implies the last limit is equal to $f(a)$.