1.10

Part a.) Choose an arbitrary $x \in [0,1]$. Clearly, the sequence $\lbrace f_n(x) \rbrace$ is monotonic-decreasing, and bounded below by $0$. Thus, $\lim_{n \to \infty} f_n(x)$ exists $\forall x \in [0,1]$. Simply define $f$ to be the point-wise limit of $f_n$.

Part b.) Recall from the construction of $\hat{\mathcal{C}}$ that: $$[0,1] \cap \hat{\mathcal{C}}^c = \bigcup_{k=1}^{\infty} \bigcup_{n = 1}^{3^{k-1}-1} O_n^k$$ Where the length of the open intervals $O_n^k$ depend on how we construct our specific $\hat{\mathcal{C}}$. Let's choose $x \in \hat{\mathcal{C}}$ arbitrarily. Define: $$S = \lbrace x \in [0,1] | x = \frac{b_{n,k} - a_{n,k}}{2}, O_n^k = (a_{n,k}, b_{n,k}) \rbrace$$ Notice, much like what we observed in (1.9), $\forall r > 0$, $B_r(x) \cap S \neq \emptyset$. Construct a sequence from $S$ in this fashion. (I.e. $\forall n$ choose any $x_n \in B_{\frac{1}{n}}(x) \cap S$.)

Certainly, we observe $x_n \to x$ as $n \to \infty$. However, by construction of $f$, notice $f(x_n) = 0$ for every $n$. Thus, $\lim_{n \to \infty} f(x_n) = 0 \neq f(x) = 1.$ Therefore $f$ must be discontinuous $\forall x \in \hat{\mathcal{C}}$.

Finally, the author wants you to observe that since $f_n$ is both Riemann integrable $\forall n$, and decreasing, that the sequence $ \lbrace \int f_n \rbrace$ is both decreasing and bounded below (by, say, $0$). Therefore $\lim_{n \to \infty} \int f_n$ exists. However, we have $\int \lim_{n \to \infty} f_n = \int f$, but $f$ clearly isn't Reimann integrable, since it's discontinuous on $\hat{\mathcal{C}}$, a set with positive measure. Thus, the point-wise limit of a sequence of Riemann-integrable functions isn't necessarily Riemann integrable in general.