Week 3 Worksheet Solutions

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1. Certainly $\mu(\emptyset) = 0 = \nu(\emptyset)$ so $\emptyset$ 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 \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) = \sum_{n=1}^\infty \mu(B_n) = \sum_{n=1}^\infty \nu(B_n) = \nu \left( \, \bigcup_{n=1}^\infty B_n \right) \] 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 $\R$ that agree on every interval $(a,b) \subset \R$. Since the collection \[ \{ (a,b) : a \l b \} \cup \{ \emptyset \} \] is a π-system (it is non-empty and closed under finite intersections) the 𝜆-system \[ \{ B \in \borel(\R) : \mu(B) = \nu(B) \} \] 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 $\R$.
3. 1. Fix $S \subset \R$. We must prove that \[ \Lambda(S) \ge \Lambda(S \cap F) + \Lambda(S \cap F^{\mathsf{c}}) \] holds. First note that \[ \Lambda(S \cap F) \le \Lambda(S \cap E) = 0 \] gives $\Lambda(S \cap F) = 0$. Since $F \subset E$ we have \[ \Lambda(S \cap F^{\mathsf{c}}) \ge \Lambda(S \cap E^{\mathsf{c}}) \] so \[ \begin{aligned} \Lambda(S) & {} = \Lambda(S \cap E) + \Lambda(S \cap E^{\mathsf{c}}) \\ & {} \ge \Lambda(S \cap E^\mathsf{c}) \\ & {} \ge \Lambda(S \cap F^\mathsf{c}) \\ & {} = \Lambda(S \cap F) + \Lambda(S \cap E^\mathsf{c}) \end{aligned} \] as desired.
   2. We have seen in the first tutorial that the middle-thirds Cantor set $\mathcal{C}$ has $\Lambda(\mathcal{C}) = 0$. Since it is closed it is a Borel subset of $\R$ and therefore belongs to the Lebesgue σ-algebra on $\R$.
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 \[ \int \lim_{n \to \infty} g_n \intd \mu = \lim_{n \to \infty} \int g_n \intd \mu \] which says \[ \int f_1 \intd \mu - \int \lim_{n \to \infty} f_n \intd \mu = \int f_1 \intd \mu - \lim_{n \to \infty} \int f_n \intd \mu \] and can be rearranged to give \[ \lim_{n \to \infty} \int f_n \intd \mu = \int \lim_{n \to \infty} \int f_n \intd \mu \] because the integral of $f_1$ is finite.
   
   The assumption is necessary as otherwise the result is false with $X = \N$ and $\B = \mathcal{P}(\N)$ and $\mu$ counting measure and $f_n$ the indicator function of $\{ j \in \N : j \ge n \}$.
5. Write \[ \{ x \in X : f(x) > 0 \} = \bigcup_{n \in \N} \{ x \in X : f(x) \ge \tfrac{1}{n} \} \] and note that $E(n) = f^{-1}([\tfrac{1}{n},\infty))$ is measurable for every $n \in \N$ because $f$ is measurable. It suffices to prove for every $n \in \N$ that $\mu(E(n)) = 0$ because of subadditivity. Suppose there is $n \in \N$ with $\mu(E(n)) > 0$. Since $f \ge \tfrac{1}{n} 1_{E(n)}$ we must have \[ \int f \intd \mu \ge \int \dfrac{1}{n} 1_{E(n)} \intd \mu > 0 \] which is a contradiction.
6. Write \[ f_n(x) = 1_{[0,n]} \cdot \left( 1 - \dfrac{x}{n} \right)^n e^{x/2} \] for all $n \in \N$ and all $x \in \R$. The sequence $n \mapsto f_n(x)$ is non-decreasing for every $x \in \R$. The pointwise limit of the sequence $f_n$ is \[ g(x) = 1_{[0,\infty)}(x) \cdot e^{-x/2} \] which has integral 1/2 by Riemann sum considerations. By the monotone convergence theorem \[ \lim_{n \to \infty} \int f_n \intd \mu = \dfrac{1}{2} \] as desired.
7. Yes. $1_E f$ is measurable whenever $E \in \B$ the mapping $\nu$ is well-defined. Since $1_\emptyset = 0$ the function $1_\emptyset f$ is zero everywhere and $\nu(\emptyset) = 0$. Fix a sequence $E(1),E(2),\dots$ of pairwise disjoint sets in $\B$. Write $F$ for the union of the sequence. Certainly \[ 1_F f = \sum_{n=1}^\infty 1_{E(n)} f \] and the monotone convergence theorem together with linearity of the integral gives \[ \nu(F) = \sum_{n=1}^\infty \int 1_{E(n)} f \intd \mu = \sum_{n=1}^\infty \nu(E(n)) \] so $\nu$ is a measure.
8. Fix a sequence of functions $f_n : \N \to [0,\infty]$. The monotone convergence theorem gives \[ \int \sum_{n=1}^\infty f_n \intd \mu = \sum_{n=1}^\infty \int f_n \intd \mu \] which, for counting measure, becomes \[ \sum_{j=1}^\infty \sum_{n=1}^\infty f_n(j) = \sum_{n=1}^\infty \sum_{j=1}^\infty f_n(j) \] 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 \[ \lim_{n \to \infty} \int g_n \intd \mu = \int \lim_{n \to \infty} g_n \intd \mu = \int \liminf_{n \to \infty} f_n \intd \mu \] and since $g_n \le f_n$ for all $n \in \N$ we have \[ \lim_{n \to \infty} \int g_n \intd \mu \le \liminf_{n \to \infty} \int f_n \intd \mu \] as desired.
10. 1. Since $0 \le f_n(x) \le \tfrac{1}{n}$ for all $n \in \N$ and all $x \in \R$ the pointwise limit is zero. As $f_n$ is simple we may immediately calculate that its integral is 1 for all $n \in \N$. 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 $(\tfrac{1}{n},\tfrac{2}{n}]$ for all $n \in \N$ the pointwise limit is zero. As $f_n$ is simple we may immediately calculate that its integral is 1 for all $n \in \N$. 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 $(X,\B,\mu)$ and a non-decreasing sequence $n \mapsto f_n$ of measurable functions from $X$ to $\R$. Fatou's lemma gives \[ \int \lim_{n \to \infty} f_n \intd \mu \le \lim_{n \to \infty} \int f_n \intd \mu \] which is the difficult direction in the proof of the monotone convergence theorem. The other direction follows from monotonicity of the integral.