I did the following proof,is it correct or do I need to do slight changes?
Theorem
Let, $f:[a,b]\to \mathbb R$ be a function of bounded variation on $[a,b]$ and $\phi:[a,b]\to \mathbb R$ be a function on $[a,b]$ such that $\phi(x)=f(x)$ except at one point,then $\phi$ is of bounded variation on $[a,b]$.
Proof: Let $\phi(x)=f(x)$ ,except at $c\in [a,b]$.
Since $f$ is bounded,so is $\phi$ on $[a,b]$,and $|\phi(x)|\leq M$ for all $x\in [a,b]$ ( for some $M>0$)
Let $P$ be a partition of $[a,b]$ of the form $\{a=x_0<x_1<...<x_n=b\}$.
If $c\notin P$,then we are done as $V(\phi,P)=V(f,P)\leq V_a^b(f)<V_a^b(f)+2M$.
If $c\in P$,then $c=x_k$ for some $k=1,2,...,n$
Then,$V(\phi,P)=\sum\limits_{i=1}^n|\phi(x_{i-1})-\phi(x_i)|=|\phi(x_{k-1})-\phi(x_k)|+\sum\limits_{i=1}^{k-1}|\phi(x_{i-1})-\phi(x_i)|+\sum\limits_{i=k+1}^n|\phi(x_{i-1})-\phi(x_i)|$$=|\phi(x_{k-1})-\phi(x_k)|+\sum\limits_{i=1}^{k-1}|f(x_{i-1})-f(x_i)|+\sum\limits_{i=k+1}^n|f(x_{i-1})-f(x_i)|\leq2M+V_a^b(f)$
So,$V(\phi,P)\leq 2M+V_a^b(f)$ for each $P\in \mathcal P[a,b]$.
So,$\phi$ is of bounded variation on $[a,b]$.