We are are asked to find an example of a measure $\lambda$ such that $$f(x)=\int_0^x d\lambda >0 $$ for all $x>0$, $f$ is continuous in $[0,1]$, and $\lambda$ is mutually singular wrt the Lebesgue measure $m$ in $[0,1]$.
After some searching, we saw that we can construct a measure which is concentrated in the Cantor set $C$. Namely, we define $\lambda(I)=m(c(I\cap C))$, where $c$ is the Cantor function. In that way, we cleary have the "mutually singular" part. We have more trouble proving that $f>0$ and $f$ is continuous in $[0,1$]. Our problem is that we think we need to prove that $f$ is actually the Cantor function $c$, but we don't know how to do it (is this correct?) Any help will be appreciated, thank you.
I think you are looking at this somewhat backwards. You want to find a positive Borel measure supported on the Cantor set $C$, such that $\lambda(C)=1$. Since $m(C)=0$, $\lambda$ would be mutually singular to $m$. Also, $m$ of a number is not defined.
Essentially, I think you are looking for the $\lambda$ satisfying for $f(x)=c(x)$ that $c(x)=\int_{0}^x d \lambda$. Alternatively, $\lambda\big( [0,x] \big)=c(x)$. I think the standard way to this is using the Lebesgue-Stieltjes measure or the Carathéodory's extension theorem. i.e., the measure obtained from the premeasure $\lambda_0\big((a,b]\big)= c(b)-c(a)$. With this approach, you just need to show/know that $c(x)$ is continuous.