I was looking to clarify the last step in Rudin's proof of the Riesz Representation theorem. The last step has us prove that:
For every $f \in C_c (X)$, $\Lambda f=\int_X f d\mu$.
Now there are a few steps toward the end that didn't quite register with me.
1) We have that there are functions $h_i \prec V_i$ such that $\sum h_i=1$ on K. So, $f=\sum h_i f$. This is clear, but the next part is where I'm not certain:
Step II claims: If $K$ is compact, then $K\in M_f$ and $$\mu (K)=inf\{\Lambda f:K \prec f\}$$
Now this is supposed to be used to show that:
$$\mu (K)\leq \Lambda (\sum h_i)= \sum \Lambda h_i$$
I was trying to work this out and have the following:
Since $K\subset \cup V_i $ for $i=1,…,n$, we have that $\mu (K)= \Lambda_{inf K\prec h_i} h_i$, so then certainly,
$$\mu(K)\leq \sum \Lambda h_i$$
However, I'm not sure if this is the right way to go about this.
2) The final part of this step is on page 47. After following the necessary deductions, we arrive to:
$$\sum^{n}_{i=1} (|a|+y_i +\epsilon)[\mu(E_i)+\frac{\epsilon}{n}]-|a|\mu(K)
\\=\sum^{n}_{i=1} (y_i -\epsilon)\mu(E_i)+2\epsilon \mu(K) + \frac{\epsilon}{n}\sum^{n}_{i=1} (|a|+y_i + \epsilon)
\\ \leq \int_X fd\mu + \epsilon[2\mu(K) +|a|+b +\epsilon]$$
I am having trouble understanding how he got to the last two steps (the first equality and then the second inequality).
Best Answer
To answer your first question, we can use similar logic to what was used in Step II. Let $$ 0\lt\alpha \lt 1$$ and $$W = \{x:\sum h_i > \alpha\}$$ Then $K\subset W$ (since $\sum h_i = 1 \;on\; K$) and W is open. Also if $g$ is a function such that $g\prec W$, then $g\le\frac{1}{\alpha}\sum h_i$. Thus, $$\mu(K)\le\mu(W)=sup\{\land g:g\prec W\}\le\land(\frac{1}{\alpha}\sum h_i)=\frac{1}{\alpha}\sum \land h_i$$Since this is true for every $\alpha \lt 1$, $\mu(K) \le \sum \land h_i$.
For the second question, $$\sum_{i=1}^n(\lvert a \rvert+y_i +\epsilon)[\mu(E_i)+\frac{\epsilon}{n}]-\lvert a \rvert\mu(K)=\sum_{i=1}^n(\lvert a \rvert+y_i +\epsilon)\mu(E_i)+\frac{\epsilon}{n}\sum_{i=1}^n(\lvert a \rvert+y_i +\epsilon)-\lvert a \rvert\mu(K)=\sum_{i=1}^n\lvert a \rvert\mu(E_i)+\sum_{i=1}^n(y_i +\epsilon)\mu(E_i)+\frac{\epsilon}{n}\sum_{i=1}^n(\lvert a \rvert+y_i +\epsilon)-\lvert a \rvert\mu(K)$$
Now because $E_i$ are all disjoint and their union is $K$, we have $\sum_{i=1}^n\lvert a \rvert\mu(E_i) = \lvert a \rvert\mu(K)$ so the $\lvert a \rvert\mu(K)$ terms cancel above. Also, $$\sum_{i=1}^n(y_i +\epsilon)\mu(E_i)=\sum_{i=1}^n(y_i -\epsilon+2\epsilon)\mu(E_i)=\sum_{i=1}^n(y_i -\epsilon)\mu(E_i)+\sum_{i=1}^n2\epsilon\mu(E_i)=\sum_{i=1}^n(y_i -\epsilon)\mu(E_i)+2\epsilon\mu(K)$$ Putting all of these steps together gives you the first equality. For the inequality, note that for every $i=1,2,...n$, $y_i-\epsilon \lt f(x)$ for every x in $E_i$. Hence, $$\sum_{i=1}^n(y_i -\epsilon)\mu(E_i) \le \int_X fd\mu$$. Also note that for every $i=1,2,...n$, $y_i\le b$ so $\sum_{i=1}^n(\lvert a \rvert+y_i +\epsilon) \le n(\lvert a \rvert+b +\epsilon$. Putting everything together gives you the inequality.