It is possible to give a counterexample without having to construct it explicitly. It suffices to prove this for real continuous functions, since the definition of integration of complex functions and the linearity of the limit allow the extension of the result to all members of $C[0,1]$. Let $I_{n,j}= [\frac jn, \frac {j+2^{-n}}{n}]$ and define continuous positive functions $\phi_{n,j}$ on $I_{n,j}$ such that $\int_{I_{n,j}} \phi_{n,j} = \frac 1n$ and put $h_n= \sum_{j=0}^{n-1}\phi_{n,j}$ Notice $m\{x:h_n(x)\ne 0\}=\sum m(I_{n,j})=n\frac{2^{-n}}{n}=2^{-n} \to 0 \ \text{as} \ n \to \infty$. So that $h_n$ are non-negative continuous functions that converge to $0$ almost everywhere on $[0,1]$.
For any real continuous hence uniformly continuous $f$ on the compact unit interval, and any $j=0,...,n-1$ we have that $|f(x)-f(y)|<\epsilon$ with $x,y \in [\frac jn, \frac {j+1}n]$ (having chosen a sufficiently large $n$, of course).
By the above $f(\frac jn) -\epsilon
$f(\frac jn)\phi_{n,j} -\epsilon\phi_{n,j}-f(\frac jn)\le f(x)\phi_{n,j}-f(\frac jn)\le\epsilon\phi_{n,j}+f(\frac jn)\phi_{n,j}-f(\frac jn)$ Hence $\int_{\frac jn}^{\frac {j+1}n}f(\frac jn)\phi_{n,j} -\epsilon\phi_{n,j}-f(\frac jn)\le \int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}-f(\frac jn)\le\int_{\frac jn}^{\frac {j+1}n}\epsilon\phi_{n,j}+f(\frac jn)\phi_{n,j}-f(\frac jn)$ Which implies $\frac 1n \left(f(\frac jn) -\epsilon-f(\frac jn)\right)\le \int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}-f(\frac jn)\le\frac 1n \left(f(\frac jn) +\epsilon-f(\frac jn)\right)$ Summing over $j$ and noting that $\int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}=\int_0^1f(x)\phi_{n,j}$ the above gives $-\epsilon \le \sum \int_0^1f(x)\phi_{n,j}-f(\frac jn) dx \le \epsilon$ thus $\left | \int_0^1 f(x)\sum \phi_{n,j}-\frac 1n \sum f(\frac jn)\right|\le \epsilon$