Integrals Relative to Lebesgue Measure

(SC4b) If $f(\boldsymbol {x})=-\infty$, then $f$ is USC at $\boldsymbol {x}$ iff $\lim_{\boldsymbol {y} \rightarrow \boldsymbol {x}} f(\boldsymbol {y})= -\infty$. (SC7b) $\bar{f}$ is USC, and is the minimal USC function $\geqslant f$. (SC9) Let $A$ denote a generic nonempty index set. For each $\alpha \in A$, suppose that $f_{\alpha }$ is a function from $\mathbb {R}^{k}$ into $\overline{\mathbb {R}}$. We have the following:

Let $E \subset \overline{\mathbb {R}}^{k}$.

Let $K$ denote a compact subset of $\mathbb {R}^{k}$.

Let $A$ denote a generic nonempty index set, and suppose for each $\alpha \in A$ that $f_{\alpha }$ is continuous.

For each $\boldsymbol {x} \in \mathbb {R}^{k}$ and $\delta >0$, define $\phi (\boldsymbol {x})=\inf_{\delta >0} \sup_{\boldsymbol {s}, \boldsymbol {t} \in B(\boldsymbol {x}, \delta )}\left|f(\boldsymbol {s})-f(\boldsymbol {t})\right|$.

Complete the proof of (SC4).

(SC4): We have the following: (a) If $f(\boldsymbol {x})=+\infty$, then $f$ is LSC at $\boldsymbol {x}$ iff $\lim_{\boldsymbol {y} \rightarrow \boldsymbol {x}} f(\boldsymbol {y})=+\infty$. (b) If $f(\boldsymbol {x})=-\infty$, then $f$ is USC at $\boldsymbol {x}$ iff $\lim_{\boldsymbol {y} \rightarrow \boldsymbol {x}} f(\boldsymbol {y})=-\infty$.

Give a formal writeup of the proof of the following [intuitively obvious] assertion: the set of $\boldsymbol {x} \in I$ such that $\boldsymbol {x}$ is a boundary point of some subrectangle of $I$ at some stage is $\lambda_{k}$-null. This assertion was made at the end of the proof of (SC3).

(SC3): If $f(\boldsymbol {x})=-\infty$, then $f$ is LSC at $\boldsymbol {x}$. If $f(\boldsymbol {x})=+\infty$, then $f$ is USC at $\boldsymbol {x}$. [Vacuously true.]

$\mathcal{B}_{I}^{k}$ is the $\sigma$-field on $I$ generated by the $k$-dimensional right semiclosed subrectangles $(\boldsymbol {c}, \boldsymbol {d}]$ of $I$, and $\mathcal{L}_{I}^{k}=\overline{\mathcal{B}_{I}^{k}}$, where the completion is with respect to $\lambda_{k}|_{\mathcal{B}_{I}^{k}}$. This exercise justifies using the notation we have used.

Focus attention on a closed and bounded $k$-dimensional rectangle $I=[\boldsymbol {a}, \boldsymbol {b}]$. Suppose that $f$ and $g$ are RI on $I$. Using the main results of this section, quickly show that $af+bg$ [where $a, b \in \mathbb {R}$] and $fg$ are also RI. [The idea is to not regurgitate the calculus-based proofs. Observe by Claim 1 or Claim 2 (of this section) that $af+bg$ and $fg$ are continuous almost everywhere on $I$ if $f$ and $g$ are RI.]

Here, take $I=[0,1]$, and let $f$ denote the indicator function of the set of rational numbers in $[0,1]$.

Work the following parts:

Obtain $\int_{(0,1)} \frac{1}{\sqrt{x}} \, d\lambda (x)$ if it exists, otherwise show that it fails to exist. Observe that $\frac{1}{\sqrt{x}}$ is not bounded on $(0,1)$, so that one cannot immediately appeal to Claim 1 or Claim 2 (of this section).

$\int_{\mathbb {R}} \frac{1}{1+x^{2}} \, d\lambda (x)$ exists and equals $\pi$. [Be more careful here than merely stating that the improper Riemann integral $\int_{-\infty }^{+\infty } \frac{1}{1+x^{2}} \, dx$ exists and equals $\pi$.]

Is a RI function necessarily Borel measurable?

Show that $\int_{1}^{+\infty } \frac{1}{x} \, d\lambda (x)=+\infty$. [Be careful here.]

Let $f: \mathbb {R} \rightarrow \overline{\mathbb {R}}$ be such that $\int_{\mathbb {R}} f \, d\bar{\lambda }$ exists, is finite. Further suppose that $\int_{(a,b)} f \, d\bar{\lambda }=0$ for every $-\infty <a<b<+\infty$. Then $f=0$ $\bar{\lambda }$-a.e. on $\mathbb {R}$. [Work from indicators to simple functions, then to nonnegative functions, then to general integrable functions.]

Assume the result of Exercise 8 [problem 7.4.8 above]. Show that there cannot exist a set $E \in \mathcal{B}$ such that $\lambda \left(E \cap [a, b]\right)=\frac{1}{2}(b-a)$ for all real $a<b$. [Write the left side as $\int_{[a,b]} \; \mathbb {1}_{E} \, d\lambda$, and write the right side as $\int_{[a,b]} \frac{1}{2} \, d\lambda$. Use the previous exercise, working by contradiction.]

There exists a real valued $f$ on $[0,1]$ such that $f$ is continuous on $(0,1]$, $\lim_{\delta \rightarrow 0^{+}} \int_{[\delta , 1]} f \, d\bar{\lambda }$ exists and is finite, but $\int_{[0,1]} f \, d\bar{\lambda }$ is undefined.

Suppose that $f$ is improperly Riemann integrable over the real line, so that

exists as a real number. [Observe that this implies that $f$ is RI over each interval $[a, b]$.] Give sufficient conditions for $f$ so that $\int_{\mathbb {R}} f \, d\lambda =\int_{-\infty }^{+\infty } f(x) \, dx$.

For general elements $f \in L^{1}$, define $\left\| f\right\|_{1}=\int \left|f\right| \, d\bar{\lambda }_{k}$. Then

For elements $f, g \in L^{1}$ write $f \sim g$ iff $f=g$ $\bar{\lambda }_{k}$-a.e. Then $\sim$ is an equivalence relation.

For $f, g \in L^{1}$, define the $L^{1}$-distance between $f$ and $g$ to be $\left\| f-g\right\|_{1}$. Show that this distance function is a metric if we agree to identify functions that are equal almost everywhere.

Here we will show that the collection $C_{c}$ of continuous functions with compact support is dense in $L^{1}$, where the notion of distance is as given in the previous exercise. That is, given $f \in L^{1}$ and $\varepsilon >0$, there is $g \in C_{c}$ such that $\left\| f-g\right\|_{1}<\varepsilon$. Work in the following stages.

We know that if $f \in L^{1}$, then $\tau_{\boldsymbol {y}} f \in L^{1}$ for every $\boldsymbol {y} \in \mathbb {R}^{k}$. [$\tau_{\boldsymbol {y}}$ is the translation operator given in Claim 2.] Show that $\lim_{\boldsymbol {y} \rightarrow \boldsymbol {0}} \left\| \tau_{\boldsymbol {y}} f-f\right\|_{1}=0$.