Can I interchange the limit and integral of a sequence of functions which is not uniformly convergent in $[0,1]$ i.e $f_n \not\to f$ uniformly is it true that $\int_0^{x_n}f_n \to \int_0^1f$ for $x_n\to 1$?
My guess is no, just thinking in terms of a picture. But if we consider $f_n(x)=x^n$ on $[0,1]$. Each function $f_n(x)$ is continuous, but the limit function $f(x)$ is not continuous: $$ f(x)=\left\{ \begin{array}{ll} 0, 0\leq x<1\\ 1, x=1\\ \end{array} \right. $$. Hence it is not uniformly convergent. But here $\int_0^{x_n}f \to \int_0^1f$ for $x_n\to 1$
Can anyone help me with one counterexample or prove it if the statement is true?
If you have a sequence of continuous functions that don't converge uniformly, then they might not converge to a continuous function, and depending on what definition of integration you are using, that might be a problem. However, limits of measurable functions are measurable, so if you are doing measure theory, things are okay.
Limits commute with integrals in nice cases. For example, there is the monotone convergence theorem and the dominated convergence theorem, and there is a discussion of why the conditions are actually necessary on those pages. But you may want to take those theorems as hints: since they are true, they tell you something about what a sequence of functions must look like to produce a counterexample.
But I will give the following partial hint to keep you from looking at those pages: You can find an example of continuous functions that converge pointwise to 0 on $[0,1]$ but every integral is equal to $1$.