2.15

It is clear by direct calculation (or from Problem 10, Part II) that our function:

$f(x) = \Bigg\lbrace \begin{array}{cc} x^{-1 / 2} & x \in (0,1) & \\ 0 & \text{otherwise} \\ \end{array}$

...is $L^1(\mathbb{R})$.

The goal of the exercise is to prove that for some fixed enumeration of the rationals $\lbrace q_n \rbrace_{n=1}^\infty$, $$F(x) = \sum_{k=1}^\infty 2^{-k} f(x - q_k)$$ ...is also $L^1(\mathbb{R})$. Proceed by simply defining the sequence of functions: $$F_n(x) = \sum_{k=1}^n 2^{-k} f(x - q_k)$$ ...and first observing that $F_n(x)$ is $L^1(\mathbb{R}) \hspace{0.25cm} \forall n$: $$\int_\mathbb{R} F_n(x) dx = \int_\mathbb{R} \sum_{k=1}^n 2^{-k} f(x - q_k) dx = \dots$$ $$\ldots = \sum_{k=1}^n 2^{-k} \int_\mathbb{R} f(x - q_k)dx = C \sum_{k=1}^n 2^{-k}$$ Where $C = \int_\mathbb{R} f(x)dx$ (by translation invariance). Thus, since $\lbrace F_n \rbrace_{n=1}^\infty$ is a collection of non-negative measurable functions such that $F_n \leq F_{n+1} \hspace{0.25cm} \forall n$, and $F_n \nearrow F$ monotonically, we have by the Monotone Convergence Theorem, $$\int_\mathbb{R} |F(x)| dx = \int_\mathbb{R} F(x)dx = \lim_{n \to \infty} \int_\mathbb{R} F_n(x)dx = \ldots$$ $$\ldots = C \sum_{k=1}^\infty 2^{-k} = C < \infty$$ Thus, $F \in L^1(\mathbb{R})$. Next, observe that even if some function $G = F$ almost everywhere, on any fixed interval $I \subset \mathbb{R}, G|_I = F|_I$ almost everywhere. Thus, the sets: $$\mathcal{G}_I^n = \lbrace x \in G|_I \hspace{0.25cm} \big| \hspace{0.25cm} G(x) > n \rbrace \hspace{0.25cm} \text{and} \hspace{0.25cm} \mathcal{F}_I^n = \lbrace x \in F|_I \hspace{0.25cm} \big| \hspace{0.25cm} F(x) > n \rbrace$$ Have the same (positive $\forall n$) measure. Thus, $G$ must be unbounded on any interval.