An alternative proof for Riesz Representation Theorem for $C([0,1])$ During my class in real analysis, my teacher mentioned an alternative method to prove Riesz's theorem for $C([0,1])$. He just mentioned the method, but did not prove it. (A usual way of doing this is provided here: Riesz representation theorem for $C([0,1])$)

Step 1: Let $\Lambda$ be a nonnegative linear functional on $C[0,1]$, and define
$$F(t)=\inf\left\{\Lambda(f):f\ge\textbf{1}_{[0,t]}\right\}$$ $F(t)$
is nondecreasing function of $t$,  and $F(t)$ is right continuous.
Step 2: Let $\mu$ be the measure on $[0,1]$ s.t. $\mu([0,t])=F(t)$. Show that $$\Lambda(f)=\int fd\mu$$ for all $f\in C[0,1]$.

I am trying to prove this using his method, but I get stuck for a while and do not have much progress. Here are my progress so far:
First, I try to show $F(t)$ is a nondecreasing function. Let $0<t_1<t_2\leq 1$, since $\Lambda(f_i)\leq F(t_i)+\varepsilon$ for $i=1,2$, then $f=f_1+f_2\ge\textbf{1}_{[0,t_1]\cup[0,t_2]}$ and $\Lambda(f)\leq F(t_1)+F(t_2)+2\varepsilon$.
My aim is to show that $F(t_1)\leq F(t_2)$, but I cannot get a lower bound for $\Lambda(f)$. If I should use Urysohn lemma, how should I find the two disjoint sets to make separations?
Added: For any $0<t_1<t_2\leq 1$, observe that
$$\{\Lambda(f):f\ge\textbf{1}_{[0,t_2]}\}\subset\{\Lambda(f):f\ge\textbf{1}_{[0,t_1]}\}$$

Take infimum on both sides, $F(t_1)\leq F(t_2)$.
Second, I try to show $F(t)$ is right continuous, i.e. $$\lim_{n\to\infty}F\left(t+\frac{1}{n}\right)=F(t)$$
I proceed this way:
$$\lim_{n\to\infty}F\left(t+\frac{1}{n}\right)=\lim_{n\to\infty}\inf\left\{\Lambda(f):f\ge\textbf{1}_{[0,t+\frac{1}{n}]}\right\}$$
From here, I'd like to take intersection over $n\ge 1$, but I'm not sure whether I can do this to get $F(t)$.
Added: For fixed $t\in(0,1]$ and any $\varepsilon>0$, choose $f_1(x)\ge\textbf{1}_{[0,t]}$ s.t. $\Lambda(f_1)\leq F(t)+\varepsilon$. Then define $f_2(x)=f_1(x)+\varepsilon$, then $f_2(t)\ge 1+\varepsilon$. Choose $N\in\mathbb{N}$ large s.t. for all $n>N$, $f_2(t+\frac{1}{n})>1$ by continuity of $f_2(x)$. Hence $f_2(x)\ge\textbf{1}_{[0,t+\frac{1}{n}]}$. Plug this result into $\Lambda(\cdot)$ and take infimum on both sides, we get the estimate:
$$F\left(t+\frac{1}{n}\right)\leq\Lambda(f_2)=\Lambda(f_2-f_1+f_1)=\varepsilon+\Lambda(f_1)\leq F(t)+2\varepsilon$$
Let $n\to\infty$ and $\varepsilon\to 0$, we prove right continuity:
$$\lim_{n\to\infty}F\left(t+\frac{1}{n}\right)=F(t)$$
For Step 2, though I do not know how to write this in detail currently, I think the process should be similar to the method provided in the post I mentioned. The key in this step is to partition $[0,1]$ into closed sets $[a_i,b_i]$ of size less than $\delta$, and to define $f_1(x)$ and $f_2(x)$ (like in the post I mentioned) to approximate $f(x)$, and then finish the argument by reaching a bound for $\left|\Lambda(f)-\int_{0}^{1}f(x)d\mu\right|$, where the bound should consist of $\|\Lambda(f)\|$ and the measure $\mu$. But I really do not know how to reach this step.
Added: By definition of measure $\mu$ on $[0,1]$, $\mu([0,1])=F(1)=\inf\{\Lambda(f):f\ge\textbf{1}_{[0,1]}\}=\Lambda(\textbf{1}_{[0,1]})$. WLOG, we can assume that $\Lambda(\textbf{1})=1$. Hence, $\mu([0,1])=1$. Since $f\in C([0,1])$, it is bounded. WLOG, we can assume $0\leq f(x)\leq 1$.
To show $\ge$, if we change $f$ to $1-f$ in the inequality, and since $\Lambda(\textbf{1})=\mu([0,1])$ as we have just assumed, we achieve equality. So this side is trivial.
To show $\leq$, let $\{[a_j,b_j]\}_{j=1}^{n}\subset [0,1]$ be a finite disjoint collection with $b_j-a_j\leq\varepsilon$, and $\mu\{x: f(x)\notin\bigcup_{j=1}^{n}[a_j,b_j]\}\leq\varepsilon$. Define $E_j=\{x:f(x)\in[a_j,b_j]\}$. Note that $\{E_j\}$'s are disjoint. Hence by Urysohn's lemma, $\exists\{g_j(x)\}\in[0,1]$ continuous s.t. for every $j=1,\ldots,n$,
$$g_j(x)=\begin{cases}1, &x\in E_j\\0, &x\notin E_j\end{cases}$$
Then by partition of unity, define $h_1=g_1$, $h_j=(1-g_1)\ldots(1-g_{j-1})g_j$ for $2\leq j\leq n-1$, and $h_n=(1-g_1)\ldots(1-g_{n-1})$. Then $\sum_{j=1}^{n}h_j(x)=1$ and $h_j(x)=1$ on $E_j$. Then $f=\sum_{j=1}^{n}f\cdot h_j$, where $f\cdot h_j\ge a_j\textbf{1}_{E_j}$. Using this estimate, we can show $\leq$:
$$\int_{0}^{1}fd\mu\leq\sum_{j=1}^{n}b_j\mu(E_j)+\varepsilon\leq\sum_{j=1}^{n}(a_j+\varepsilon)\mu(E_j)+\varepsilon\leq\sum_{j=1}^{n}\Lambda(f\cdot h_j)+2\varepsilon=\Lambda(f)+2\varepsilon$$
Let $\varepsilon\to 0$, done.
Can anyone help me to complete the proof based on this method? I'll update accordingly if I have more progress. Thank you!
 A: Here is a feasible way to complete part 3. The idea is to write the integral representation as a Riemann integral (functions of BV), and use sums to approximate.
Let $\mathcal{P}=\{0=x_0<x_1<\ldots<x_n=1\}$ be a partition of $[0,1]$. By definition of measure $\mu$ s.t. $F(t)=\mu([0,t])$, we can express$$\int_{0}^{1}fd\mu=\int_{0}^{1}fdF(x)$$
We can approximate the integral on the right by sums, $$\sum_{k=0}^{n-1}f(x_{k+1})\left(F(x_{k+1})-F(x_k)\right)$$, as $|\mathcal{P}|\to 0$. Since $F(t)$ is right continuous, we need to split into two cases: $f(0)=0$ and $f(0)=a\neq 0$ for $a$ any constant. For the first case, define the step function s.t. for $|x_{k+1}-x_k|<\delta$,
$$f_n(x)=\sum_{k=0}^{n-1}f(x_k)(\textbf{1}_{(x_k,x_{k+1}]})$$
Since we have assumed $f(0)=0$, we can exclude the first term and hence $$f_n(x)=\sum_{k=0}^{n-1}f(x_{k+1})(\textbf{1}_{[0,x_{k+1}]}-\textbf{1}_{[0,x_k]})$$
By definition of $F(t)$, for any $\varepsilon>0$,
$$\left|\Lambda(f_n)-\sum_{k=0}^{n-1}f(x_{k+1})\left(F(x_{k+1})-F(x_k)\right)\right|<\varepsilon$$
Since $\sum_{k=0}^{n-1}f(x_{k+1})\left(F(x_{k+1})-F(x_k)\right)\to\int_{0}^{1}fdF(x)$ as $|\mathcal{P}|\to 0$,
$$\left|\Lambda(f_n)-\int_{0}^{1}fdF(x)\right|<\varepsilon$$
Note that for every $x\in[0,1]$, $|f_n(x)-f(x)|<\varepsilon$. Take supremum on both sides, $\|f_n-f\|_{\infty}<\varepsilon$ for any $\varepsilon>0$. Thus,
$$|\Lambda(f_n)-\Lambda(f)|\leq\|\Lambda\|\|f_n-f\|_{\infty}<\|\Lambda\|\varepsilon$$
, where $\|\Lambda\|$ is the operator norm. Note that $\|\Lambda\|<\infty$, since nonnegative linear functionals on $C[0,1]$ are continuous. By the above two inequalities,
$$\left|\Lambda(f)-\int_{0}^{1}fd\mu\right|=\left|\Lambda(f)-\int_{0}^{1}fdF(x)\right|<(\|\Lambda\|+1)\varepsilon$$
Let $\varepsilon\to 0$, done with first case.
For the second case, consider the function $g(x)=f(x)-a$ and plug in the result we yield in the first case:
$$\Lambda(f)=\Lambda(g)+a\Lambda(\textbf{1}_{[0,1]})=\int_{0}^{1}gd\mu+aF(1)=\int_{0}^{1}f d\mu$$
Done with second case.
