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The Lumenary · First-year graduate

Measure theory and integration

Name: ______________________

Date: ______________________

You will learn to

Define σ-algebras, measures and outer measures, and construct Lebesgue measure from outer measure by Carathéodory's criterion
Define the Lebesgue integral through simple functions, and prove the monotone convergence theorem, Fatou's lemma and the dominated convergence theorem
Recognise when a limit may not be exchanged with an integral, using counterexamples in which mass escapes
State and apply the Fubini–Tonelli theorem, including why σ-finiteness and integrability are needed
Prove that $L^p(\mu)$ is a Banach space (Riesz–Fischer), and relate the Lebesgue integral to the Riemann integral

1.
Let $\mathcal A$ be the collection of subsets $E \subseteq \N$ such that $E$ is finite or $\N \setminus E$ is finite. Which statement is true?
- A. $\mathcal A$ is a σ-algebra.
- B. $\mathcal A$ is closed under complements and finite unions, but not under countable unions.
- C. $\mathcal A$ is not closed under complements.
- D. $\mathcal A = \mathcal P(\N)$ .
2.
Let $f = 5 \cdot \mathbf 1_{\Q \cap [0, 1]} + 3 \cdot \mathbf 1_{[0, 2]}$ on $\R$ . Compute the Lebesgue integral $\int f \, d\lambda$ .
3.
Let $g_n = \mathbf 1_{[n, n+1]}$ on $\R$ with Lebesgue measure. Then $g_n \to 0$ pointwise, but $\int g_n = 1$ for all $n$ . Which hypothesis of the dominated convergence theorem fails?
- A.The $g_n$ are not measurable.
- B.There is no integrable $g$ with $|g_n| \le g$ for all $n$ .
- C.The convergence is only almost everywhere, not everywhere.
- D.The $g_n$ are not uniformly bounded.
4.
The proof of Fatou's lemma sets $g_k = \inf_{n \ge k} f_n$ and concludes $\int \liminf f_n = \lim_k \int g_k$ . Which result justifies this equality?
- A.The monotone convergence theorem, since $0 \le g_1 \le g_2 \le \cdots$ and $g_k \to \liminf f_n$ .
- B.The dominated convergence theorem.
- C.Fubini's theorem.
- D.Countable additivity of $\mu$ alone.
5.
For $m, n \ge 0$ let $a_{m,n} = 1$ if $m = n$ , $a_{m,n} = -1$ if $m = n + 1$ , and $a_{m,n} = 0$ otherwise. Compute $\displaystyle \sum_{m=0}^\infty \Big( \sum_{n=0}^\infty a_{m,n} \Big)$ .
6.
Which Lean statement correctly formalises Markov's inequality $\varepsilon\,\mu(\{f \ge \varepsilon\}) \le \int f \, d\mu$ for measurable $f : \alpha \to [0, \infty]$ ? (Here ℝ≥0∞ is $[0, \infty]$ and ∫⁻ is the integral of such functions.)
- A.theorem m (hf : Measurable f) (ε : ℝ≥0∞) : ε * μ {x | ε ≤ f x} ≤ ∫⁻ x, f x ∂μ
- B.theorem m (hf : Measurable f) (ε : ℝ≥0∞) : ∫⁻ x, f x ∂μ ≤ ε * μ {x | ε ≤ f x}
- C.theorem m (hf : Measurable f) (ε : ℝ≥0∞) : μ {x | ε ≤ f x} ≤ ∫⁻ x, f x ∂μ
- D.theorem m (hf : Measurable f) (ε : ℝ≥0∞) : ε * μ {x | ε ≤ f x} = ∫⁻ x, f x ∂μ
7.
Select all true statements.
- A. $\mathbf 1_{\Q \cap [0, 1]}$ is Lebesgue integrable on $[0, 1]$ but not Riemann integrable.
- B. $x^{-1/2}$ is Lebesgue integrable on $(0, 1]$ .
- C. $\dfrac{\sin x}{x}$ is Lebesgue integrable on $(0, \infty)$ .
- D.Every bounded Lebesgue measurable function on $[0, 1]$ is Riemann integrable.
8.
In the proof that $L^p$ is complete, given a Cauchy sequence $(f_n)$ we choose $n_1 < n_2 < \cdots$ with $\|f_{n_{j+1}} - f_{n_j}\|_p < 2^{-j}$ . Why is this subsequence needed?
- A.So that $G = |f_{n_1}| + \sum_j |f_{n_{j+1}} - f_{n_j}|$ has $\|G\|_p < \infty$ , which forces $G < \infty$ a.e. and gives pointwise absolute convergence.
- B.Because a Cauchy sequence in $L^p$ always converges pointwise, but only along subsequences in norm.
- C.Because Minkowski's inequality only holds for subsequences.
- D.It is not needed; the full sequence converges almost everywhere.
9.
Write union_length(intervals) that returns the Lebesgue measure of a finite union of closed intervals, given as a list of pairs (a, b) with a <= b. The intervals may overlap. (Countable subadditivity says the answer is at most the sum of the lengths; overlaps make it smaller.)
Computational — Practice

Answer key — Measure theory and integration

1.
$\mathcal A$ contains $\N$ , is closed under complements (finite $\leftrightarrow$ cofinite) and under finite unions (a finite union of finite sets is finite; if one set is cofinite, so is the union). So it is an algebra. It is not a σ-algebra: $\{0, 2, 4, \dots\} = \bigcup_n \{2n\}$ is a countable union of members of $\mathcal A$ but is neither finite nor cofinite. The function $\mu(E) = 0$ for finite $E$ , $1$ for cofinite $E$ is finitely additive on $\mathcal A$ but cannot be countably additive: $\N$ is a countable union of singletons.
2.
By linearity, $\int f \, d\lambda = 5\,\lambda(\Q \cap [0, 1]) + 3\,\lambda([0, 2]) = 5 \cdot 0 + 3 \cdot 2 = 6$ . The rationals in $[0, 1]$ form a countable union of singletons, each of measure $0$ , so they are null by countable subadditivity. The function $f$ is not Riemann integrable on $[0, 2]$ , since it is discontinuous at every point of $[0, 1]$ .
3.
A dominating $g$ must satisfy $g(x) \ge \sup_n g_n(x) = 1$ for every $x \ge 0$ , so $\int g \ge \lambda([0, \infty)) = \infty$ . No integrable dominating function exists, and the conclusion fails: $\lim \int g_n = 1 \ne 0 = \int \lim g_n$ . The mass 'escapes to infinity'. (The constant $1$ would dominate on a finite measure space, where the bounded convergence theorem applies.)
4.
As $k$ grows, $\inf_{n \ge k} f_n$ is an infimum over a smaller set, so $g_1 \le g_2 \le \cdots$ ; each $g_k \ge 0$ is measurable and $\lim_k g_k = \liminf_n f_n$ by definition. The monotone convergence theorem gives $\int \liminf f_n = \lim_k \int g_k$ . Then $g_k \le f_n$ for $n \ge k$ yields $\int g_k \le \inf_{n \ge k} \int f_n$ , and letting $k \to \infty$ gives Fatou's inequality.
5.
Row $0$ sums to $a_{0,0} = 1$ ; row $m \ge 1$ sums to $a_{m,m} + a_{m,m-1} = 1 - 1 = 0$ . So $\sum_m \sum_n a_{m,n} = 1$ . Summing the columns first, column $n$ contains $a_{n,n} = 1$ and $a_{n+1,n} = -1$ , so $\sum_n \sum_m a_{m,n} = 0$ . Fubini's theorem for counting measure on $\N \times \N$ does not apply because $\sum_{m,n} |a_{m,n}| = \infty$ : integrability is a genuine hypothesis.
6.
Option (a). Since $\varepsilon \mathbf 1_{\{f \ge \varepsilon\}} \le f$ pointwise and the integral is monotone, $\varepsilon\,\mu(\{f \ge \varepsilon\}) = \int \varepsilon \mathbf 1_{\{f \ge \varepsilon\}} \le \int f$ . In Lean: ε * μ {x | ε ≤ f x} ≤ ∫⁻ x, f x ∂μ, which is Lumenary.MeasureTheory.markov. Options (b)–(d) fail on constant functions, as the feedback shows.
7.
(a) and (b) are true; (c) and (d) are false. For (c): $\int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{x} dx \ge \frac{1}{(k+1)\pi} \int_{k\pi}^{(k+1)\pi} |\sin x| \, dx = \frac{2}{(k+1)\pi}$ , and the harmonic series diverges. So $\sin x/x \notin \mathcal L^1(0, \infty)$ , although its improper Riemann integral exists. For (d): the Dirichlet function is a counterexample; Lebesgue's criterion says a bounded function is Riemann integrable exactly when its discontinuities form a null set.
8.
Mere Cauchyness gives no pointwise control (the typewriter sequence converges in $L^1$ and nowhere pointwise). With $\|f_{n_{j+1}} - f_{n_j}\|_p < 2^{-j}$ , the function $G = |f_{n_1}| + \sum_j |f_{n_{j+1}} - f_{n_j}|$ satisfies $\|G\|_p \le \|f_{n_1}\|_p + 1$ by Minkowski and monotone convergence, so $G < \infty$ a.e. There the telescoping series converges absolutely, defining $F = \lim_k f_{n_k}$ a.e. Since $|F - f_{n_k}|^p \le (2G)^p \in L^1$ , dominated convergence gives $f_{n_k} \to F$ in $L^p$ , and a Cauchy sequence with a convergent subsequence converges.

def union_length(intervals):
    total = 0
    current = None
    for a, b in sorted(intervals):
        if current is None or a > current[1]:
            if current is not None:
                total += current[1] - current[0]
            current = [a, b]
        else:
            current[1] = max(current[1], b)
    if current is not None:
        total += current[1] - current[0]
    return total

After sorting, consecutive overlapping intervals merge into disjoint blocks, and the measure of a disjoint union is the sum of the lengths (finite additivity). The starter code returns $\sum (b - a)$ , which is only an upper bound: that inequality is subadditivity, and it is exactly the quantity minimised in the definition of Lebesgue outer measure.