(In the Riemann Sense, this is a lemma before the Fouriers-Mean-Convergence Theorem)
Suppose we have a square integrable function f:$[0,2\pi]\rightarrow \mathbf{C}$. We know that $\int_{a}^{b} f^2 dx$ is well defined. ($|f| \leq M$) Now I want to show that $\forall \epsilon$ there exists a continuous function $g$ such that $\int_{a}^{b} |f(x)-g(x)|^2 dx < \epsilon$ forall $x$.
The idea is to find a 'large' partition which contains enough points ${x_1,x_2,...,x_n}$ thus for every $c_i=\frac{x_i+x_{i+1}}{2}$ and we have $\int_{a}^{b} f(x) - \Sigma_{i=1}^n c_i \cdot [f(x_{i+1})-f(x_i)] < \epsilon/2$ and $\int_{a}^{b} f^2(x) - \Sigma_{i=1}^n c_i \cdot [f^2(x_{i+1})-f^2(x_i)] < \epsilon/2$. Then we can construct a function g which looks like a step function $\bar{g}:\bar{g}(x)=f(c_i)$ if $x \in [x_i,x_{i+1}]$. The main difference between $g$ and $\bar{g}$ is that $g$ is continuous, which means, we need to redefine the 'small' left interval part of $[x_i.x_{i+1}]$ and the small right interval part of $[x_i.x_{i+1}]$ such that the whole function $g$ is continuous. The text I read(Marsden) suggests that $[x_i,x_i+1]$ can be splitted into $[x_i,y_i][y_i,z_i][z_i,x_{i+1}]$ with $[x_i,y_i]< \frac{\epsilon}{8M^2n}$,$[z_i,x_{i+1}]< \frac{\epsilon}{8M^2n}$ and g is bounded by $M$.
My question is why do we pick the special value (motivation) $\frac{\epsilon}{8M^2n}$ and how to use such $g$ to verify that $\int_{a}^{b} |f(x)-g(x)|^2 dx < \epsilon$ forall $x$.
Since $\int_{a}^{b} |f(x)-g(x)|^2 dx=\int_{a}^{b} f^2(x)+g^2(x) -2f(x)g(x) dx$ the extension is quite complicated, I have no idea how to verify that this holds even though such $g$ has already been constructed.
I'm assuming you are working with Riemann integrable functions. Any such function $f$ is uniformly bounded on $[0,2\pi]$ by some constant $M$. You can choose the approximating function $g$ to also be bounded by $M$, and that gives you a reduction: $$ \int_{0}^{2\pi}|f-g|^{2}dx \le M\int_{0}^{2\pi}|f-g|dx. $$ Now you only have to approximate in the integral sense, which follows more easily from the definition of the Riemann integral. Let $\epsilon > 0$ be given. Then you can choose a partition $$ \mathcal{P} : a =x_0 < x_1 < x_2 < \cdots < x_n = b $$ such that the upper and lower Riemann sums differ by less that $\epsilon/2M$. If you let $x_k^{\star}=\frac{1}{2}(x_k-x_{k-1})$ and define $$ f^{\star}(x) = \sum_{k=1}^{n}f(x_k^{\star})\chi_{[x_{k-1},x_k)}(x), $$ where $\chi_{[x_{k-1},x_k]}(x)$ is $1$ on $[x_{k-1},x_{k})$ and $0$ elsewhere. The function $f^{\star}$ is bounded by $M$, where $M$ is a bound for $f$. And, $$ \int_{0}^{2\pi}f^{\star}(x)dx = \sum_{k=1}^{n}f(x_k^{\star})\Delta_{k}x. $$ is a Riemann sum. If $M_{k}$ is the supremum of $f$ on $[x_{k-1},x_k]$ and $m_{k}$ is the infimum of $f$ on $[x_{k-1},x_{k}]$, then $$ |f^{\star}(x)-f(x)| \le M_{k}-m_{k},\;\;\; x_{k-1} \le x < x_{k}. $$ Hence, $$ \int_{0}^{2\pi}|f^{\star}(x)-f(x)|dx \le \overline{S}_{\mathcal{P}}(f)-\underline{S}_{\mathcal{P}}(f) < \epsilon/4M, $$ where $\overline{S}_{\mathcal{P}}(f)$ and $\underline{S}_{\mathcal{P}}(f)$ are the upper and lower Riemann sums for $f$ over $\mathcal{P}$. Now you can modify the function $f$ at the endpoints of the intervals to instead be linear over $[x_k-\delta,x_k+\delta]$ where $\delta$ can be as small as you want. The function $f^{\star}$ and the modified function $f^{\star\star}$, which is now continuous, differ by at most $2M$ on the intervals where the function is modified and, hence, $$ \int_{0}^{2\pi}|f^{\star}-f^{\star\star}|dx = \sum_{k=1}^{n-1}\int_{x_k-\delta}^{x_k+\delta}|f^{\star}-f^{\star\star}|dx \le (n-1)(2M)\delta < \epsilon/4M $$ if $\delta$ is small enough. So, $$ \int_{0}^{2\pi}|f-f^{\star\star}|dx \le \int_{0}^{2\pi}|f-f^{\star}|dx + \int_{0}^{2\pi}|f^{\star}-f^{\star\star}|dx < \epsilon/2M \\ \int_{0}^{2\pi}|f-f^{\star\star}|^{2}dx \le 2M\int_{0}^{2\pi}|f-f^{\star\star}|dx < \epsilon. $$ The function $f^{\star\star}$ is continuous, and is bounded by $M$, where $M$ is a bound for $M$.