Week 3 Worksheet Solutions

Home | Assessment | Notes | Worksheets | Tutorials

1. Certainly $\mu (\varnothing )=0=\nu (\varnothing )$ so $\varnothing$ belongs to the given collection. Suppose that $B_1,B_2,\dots$ belong to the given collection and are pairwise disjoint. By definition of a measure we have $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (B_n)=\sum _{n=1}^{\mathrm{\infty }}\nu (B_n)=\nu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)$$ so the union of the $B_n$ belongs to the given collection. Thus the collection is a lambda system.
2. The π-𝜆 theorem states that if a 𝜆-system contains a π-system then the σ-algebra generated by the π-system is contained in the 𝜆-system. Suppose that $\mu$ and $\nu$ are Borel measure on $\mathbb{R}$ that agree on every interval $(a,b)\subset \mathbb{R}$. Since the collection $$\{(a,b):a<b\}\cup \{\varnothing \}$$ is a π-system (it is non-empty and closed under finite intersections) the 𝜆-system $$\{B\in \mathsf{Bor}(\mathbb{R}):\mu (B)=\nu (B)\}$$ contains the above π-system. By the π-𝜆 theorem the above 𝜆-system contains the σ-algebra generated by our π-system, which is the Borel σ-algebra. Thus $\mu$ and $\nu$ agree on all Borel subsets of $\mathbb{R}$.
3. 1. Fix $S\subset \mathbb{R}$. We must prove that $$\mathrm{\Lambda }(S)\ge \mathrm{\Lambda }(S\cap F)+\mathrm{\Lambda }(S\cap F^{\mathsf{c}})$$ holds. First note that $$\mathrm{\Lambda }(S\cap F)\le \mathrm{\Lambda }(S\cap E)=0$$ gives $\mathrm{\Lambda }(S\cap F)=0$. Since $F\subset E$ we have $$\mathrm{\Lambda }(S\cap F^{\mathsf{c}})\ge \mathrm{\Lambda }(S\cap E^{\mathsf{c}})$$ so $$\begin{array}{rl}\mathrm{\Lambda }(S)& =\mathrm{\Lambda }(S\cap E)+\mathrm{\Lambda }(S\cap E^{\mathsf{c}})\\ & \ge \mathrm{\Lambda }(S\cap E^{\mathsf{c}})\\ & \ge \mathrm{\Lambda }(S\cap F^{\mathsf{c}})\\ & =\mathrm{\Lambda }(S\cap F)+\mathrm{\Lambda }(S\cap E^{\mathsf{c}})\end{array}$$ as desired.
   2. We have seen in the first tutorial that the middle-thirds Cantor set $\mathcal{C}$ has $\mathrm{\Lambda }(\mathcal{C})=0$. Since it is closed it is a Borel subset of $\mathbb{R}$ and therefore belongs to the Lebesgue σ-algebra on $\mathbb{R}$.
4. Define $g_n=f_1-f_n$. The sequence $n\mapsto g_n$ is increasing and consists of non-negative functions so the monotone convergence theorem gives $$\int \underset{n\to \mathrm{\infty }}{lim}{g}_n\mathsf{d}\mu =\underset{n\to \mathrm{\infty }}{lim}\int g_n\mathsf{d}\mu$$ which says $$\int f_1\mathsf{d}\mu -\int \underset{n\to \mathrm{\infty }}{lim}{f}_n\mathsf{d}\mu =\int f_1\mathsf{d}\mu -\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu$$ and can be rearranged to give $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu =\int \underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu$$ because the integral of $f_1$ is finite.
   
   The assumption is necessary as otherwise the result is false with $X=\mathbb{N}$ and $\mathcal{B}=\mathcal{P}(\mathbb{N})$ and $\mu$ counting measure and $f_n$ the indicator function of $\{j\in \mathbb{N}:j\ge n\}$.
5. Write $$\{x\in X:f(x)>0\}=\bigcup _{n\in \mathbb{N}}\{x\in X:f(x)\ge \frac{1}{n}\}$$ and note that $E(n)=f^{-1}([\frac{1}{n},\mathrm{\infty }))$ is measurable for every $n\in \mathbb{N}$ because $f$ is measurable. It suffices to prove for every $n\in \mathbb{N}$ that $\mu (E(n))=0$ because of subadditivity. Suppose there is $n\in \mathbb{N}$ with $\mu (E(n))>0$. Since $f\ge \frac{1}{n}{1}_{E(n)}$ we must have $$\int f\mathsf{d}\mu \ge \int {\displaystyle \frac{1}{n}}{1}_{E(n)}\mathsf{d}\mu >0$$ which is a contradiction.
6. Write $$f_n(x)=1_{[0,n]}\cdot {(1-{\displaystyle \frac{x}{n}})}^n{e}^{x/2}$$ for all $n\in \mathbb{N}$ and all $x\in \mathbb{R}$. The sequence $n\mapsto f_n(x)$ is non-decreasing for every $x\in \mathbb{R}$. The pointwise limit of the sequence $f_n$ is $$g(x)=1_{[0,\mathrm{\infty })}(x)\cdot e^{-x/2}$$ which has integral 1/2 by Riemann sum considerations. By the monotone convergence theorem $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu ={\displaystyle \frac{1}{2}}$$ as desired.
7. Yes. $1_{E}f$ is measurable whenever $E\in \mathcal{B}$ the mapping $\nu$ is well-defined. Since $1_{\varnothing }=0$ the function $1_{\varnothing }f$ is zero everywhere and $\nu (\varnothing )=0$. Fix a sequence $E(1),E(2),\dots$ of pairwise disjoint sets in $\mathcal{B}$. Write $F$ for the union of the sequence. Certainly $$1_{F}f=\sum _{n=1}^{\mathrm{\infty }}{1}_{E(n)}f$$ and the monotone convergence theorem together with linearity of the integral gives $$\nu (F)=\sum _{n=1}^{\mathrm{\infty }}\int 1_{E(n)}f\mathsf{d}\mu =\sum _{n=1}^{\mathrm{\infty }}\nu (E(n))$$ so $\nu$ is a measure.
8. Fix a sequence of functions $f_n:\mathbb{N}\to [0,\mathrm{\infty }]$. The monotone convergence theorem gives $$\int \sum _{n=1}^{\mathrm{\infty }}{f}_n\mathsf{d}\mu =\sum _{n=1}^{\mathrm{\infty }}\int f_n\mathsf{d}\mu$$ which, for counting measure, becomes $$\sum _{j=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{f}_n(j)=\sum _{n=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{f}_n(j)$$ i.e. we may swap the orders of summations when dealing with non-negative valued functions. In fact, we needed this when proving the monotone convergence theorem, so the proof is circular unless we appeal to the definition in Worksheeet 3 of sums of non-negative functions without appealing to measure theory.
9. Define $g_n(x)=inf\{f_j(x):j\ge n\}$ for all $x\in X$. It is a non-decreasing sequence of non-negative functions, so we can apply the monotone convergence theorem, which states $$\underset{n\to \mathrm{\infty }}{lim}\int g_n\mathsf{d}\mu =\int \underset{n\to \mathrm{\infty }}{lim}{g}_n\mathsf{d}\mu =\int \underset{n\to \mathrm{\infty }}{lim inf}{f}_n\mathsf{d}\mu$$ and since $g_n\le f_n$ for all $n\in \mathbb{N}$ we have $$\underset{n\to \mathrm{\infty }}{lim}\int g_n\mathsf{d}\mu \le \underset{n\to \mathrm{\infty }}{lim inf}\int f_n\mathsf{d}\mu$$ as desired.
10. 1. Since $0\le f_n(x)\le \frac{1}{n}$ for all $n\in \mathbb{N}$ and all $x\in \mathbb{R}$ the pointwise limit is zero. As $f_n$ is simple we may immediately calculate that its integral is 1 for all $n\in \mathbb{N}$. In this case the limit of the integrals does not equal the integral of the limit. The monotone convergence theorem does not apply because the sequence $f_n$ of functions is not non-decreasing. Fatou's lemma does apply, and only tells us that $0\le 1$.
    2. Since $f_n=0$ outside $(\frac{1}{n},\frac{2}{n}]$ for all $n\in \mathbb{N}$ the pointwise limit is zero. As $f_n$ is simple we may immediately calculate that its integral is 1 for all $n\in \mathbb{N}$. In this case the limit of the integral does not equal the integral of the limit. The monotone convergence theorem does not apply because the sequence $f_n$ of functions is not non-decreasing. Fatou's lemma does apply, and only tells us that $0\le 1$.
11. Fix a measure space $(X,\mathcal{B},\mu )$ and a non-decreasing sequence $n\mapsto f_n$ of measurable functions from $X$ to $\mathbb{R}$. Fatou's lemma gives $$\int \underset{n\to \mathrm{\infty }}{lim}{f}_n\mathsf{d}\mu \le \underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu$$ which is the difficult direction in the proof of the monotone convergence theorem. The other direction follows from monotonicity of the integral.