proving uniform convergence for a uniformly continuous function

Question

I've been stuck on this question (2a) for about a day now!

Question

and havn't really gotten anywhere, I think that the series is that integral of a step function that approximates f ie:

$\Gamma_n = f(r/n)$ for r = 1,2,...,n

if I show that

$\Gamma_n \to f$ uniformly

then it is all fine, i can see how to do the questions once that is proved, but I don't see how to prove that;

$\|\Gamma-f\|_\infty \to 0$

where the $\|g\|_\infty$ is defined as $\{|g(x)|:x\in[0,1]\}$

UPDATE

Thanks to a hint in the comments I think I got an answer, but I'm not too sure about that very last bit

so we consider $\Gamma_n$ and define a partition $P_n=\{0=x_0<x_1<...<x_n=1\}$ such that for $i = 1,2,...,n$ and $x\in(x_{i-1},x_i)$

$\Gamma_n(x) = f(i/n)$

so for an $i$ we restrict $f$ to the interval $(x_{i-1},x_i)$ we can assume $f$ uniformly continuous

now we use this definition

for all $\epsilon_i > 0$ there exists a $\delta_i > 0$ such that if $x \in B(i/n,\delta_i ($note $\delta_i < 1/n $hence we can choose this $\delta$ as well by making $n$ large enough$)) \Rightarrow f(x) \in B(f(i/n),\epsilon_i)$

now as $\epsilon$ is in the mapped to space and we can choose it, let us set $\epsilon_i = \|f|_{(x_{i-1},x_i)}-\Gamma_n|_{(x_{i-1},x_i)}\|_\infty$

now if we let $\epsilon = max(\epsilon_i: i = 1,2,....,n)$

now we get $\|f-\Gamma_n\|_\infty \le \epsilon$

hence $\Gamma_n \to f$ uniformly

Then to finish off the the question just say how a the integral of a step function that converges uniformally to f is equal to the integral of f blah blah blah