Lebesgue Integration Basics Practice Questions, Quiz, and Step-by-Step Lesson - improve your math skills with focused questions and clear explanations.
Lebesgue Integration Basics Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice Lebesgue integration basics: measurable sets, indicator functions \(1_A\), null sets and almost everywhere reasoning, simple functions \(\sum a_i1_{A_i}\), nonnegative integrals, monotonicity, \(L^1\) integrability through \(\int |f|<\infty\), the monotone convergence theorem, Fatou's lemma, dominated convergence, and common traps involving infinite measure or null-set changes. If you need a refresher, open the lesson for mentally followable examples and quick checks.
How this Lebesgue integration basics practice works
1. Take the quiz: answer questions about indicators, null sets, simple functions, integrability, and convergence theorems.
2. Open the lesson: review the definitions, theorem hypotheses, and short examples before retrying.
3. Retry: return to the quiz and translate each problem into a measure computation, almost everywhere statement, or convergence theorem checklist.
What you will learn in the Lebesgue integration basics lesson
Indicators and null sets
Indicator rule: \(\int 1_A\,d\mu=\mu(A)\).
Null sets: changing values on a measure-zero set does not change the integral.
Almost everywhere: a property may fail on a null set and still hold a.e.
Simple functions and \(L^1\)
Simple functions: finite sums \(\sum a_i1_{A_i}\) over measurable sets.
Nonnegative integral: approximate from below by simple functions.
Integrable: \(f\in L^1\) means \(\int |f|\,d\mu<\infty\); \(L^1\) treats functions equal a.e. as the same class.
Fatou: \(\int\liminf f_n\le\liminf\int f_n\) for nonnegative \(f_n\).
Dominated convergence: a single \(g\in L^1\) with \(|f_n|\le g\) lets limits pass through integrals.
Common traps
Infinite measure: \(1_{\mathbb{R}}\) has infinite integral on \(\mathbb{R}\).
Pointwise convergence alone: not enough for dominated convergence.
Zero nonnegative integral: if \(f\ge0\) and \(\int f\,d\mu=0\), then \(f=0\) almost everywhere.
Ready to test the hypotheses?
Return to the quiz and check whether each question is about measure, almost everywhere equality, simple functions, integrability, monotone convergence, Fatou, or dominated convergence.
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Higher Analysis
Lebesgue Integration Basics Lesson
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Lesson overview
Purpose: Build a reliable first toolkit for Lebesgue integration: compute integrals of indicators and simple functions, use null sets and almost everywhere equality correctly, distinguish nonnegative integrals from finite \(L^1\) integrals, apply monotonicity and absolute integrability, and choose between monotone convergence, Fatou's lemma, and dominated convergence by checking hypotheses first.
Success criteria
Compute \(\int 1_A\,d\mu=\mu(A)\) and \(\int c\,1_A\,d\mu=c\mu(A)\).
Use countable additivity for disjoint measurable sets, set monotonicity \(A\subset B\Rightarrow\mu(A)\le\mu(B)\), and countable null sets.
Explain that a property holds almost everywhere when it fails only on a null set.
Integrate a nonnegative simple function \(\sum a_i1_{A_i}\) as \(\sum a_i\mu(A_i)\) over disjoint measurable pieces.
Know that the nonnegative integral may be \(+\infty\), and that \(\int f=0\) with \(f\ge0\) forces \(f=0\) a.e.
Use \(f\in L^1\) to mean \(\int |f|\,d\mu<\infty\), and \(f_n\to f\) in \(L^1\) to mean \(\int |f_n-f|\,d\mu\to0\).
Apply \(L^1\) facts: functions are finite a.e., sums stay integrable, \(|\int f|\le\int |f|\), and a.e.-equal functions represent the same \(L^1\) element.
Use monotone convergence for \(0\le f_n\uparrow f\).
Use Fatou's lemma and dominated convergence with the correct inequality and hypotheses.
Key vocabulary
Indicator: \(1_A(x)=1\) on \(A\) and \(0\) off \(A\).
Null set: a measurable set \(N\) with \(\mu(N)=0\).
Almost everywhere: true outside a null set.
Simple function: a measurable function with finite range, usually written as \(\sum a_i1_{A_i}\).
Nonnegative integral: the supremum of integrals of simple functions below \(f\).
\(L^1\): the space of integrable functions with \(\int |f|<\infty\).
Quick pre-check
Pre-check: What is \(\int 1_A\,d\mu\)?
Hint: The indicator counts measure, not boundary points or a maximum value.
Indicators turn set measure into integration
Learning goal: Treat \(1_A\) as the basic bridge between measure and integral, and use null sets without overthinking pointwise exceptions.
Key idea
For any measurable set \(A\), the Lebesgue integral of its indicator is its measure: \[\int 1_A\,d\mu=\mu(A).\] If \(c\ge0\), then \(\int c\,1_A\,d\mu=c\mu(A)\). When \(A\) and \(B\) are disjoint, \(1_A+1_B=1_{A\cup B}\), so integration reflects countable additivity. If \(A\subset B\), then \(1_A\le1_B\), hence \(\mu(A)\le\mu(B)\).
Examples
On \(\mathbb{R}\) with length measure, \(\int 1_{[1,3]}\,dx=2\).
A singleton has measure \(0\), so \(\int 1_{\{0\}}\,dx=0\).
The rationals in \([0,1]\) are countable, hence null.
\(\int 1_{\mathbb{R}}\,dx=+\infty\) on the whole real line.
Almost everywhere
A statement holds almost everywhere if the set where it fails has measure \(0\). Integrals of nonnegative measurable functions and \(L^1\) functions do not change when the function is altered on a null set.
Worked example
Example: What is \(\int_0^1 1_{\mathbb{Q}}(x)\,dx\)?
Inside \([0,1]\), the rationals are countable and have Lebesgue measure \(0\). Therefore \(1_{\mathbb{Q}}\) is equal to \(0\) almost everywhere on \([0,1]\), and the integral is \(0\).
Try it
Try it: If a property fails only on the rationals in \([0,1]\), does it hold almost everywhere?
Hint: A countable set has Lebesgue measure \(0\).
Simple functions are the building blocks of the integral
Learning goal: Compute simple-function integrals and see why nonnegative integrals are built by approximation from below.
Key idea
A nonnegative simple function has the form \(s=\sum_{i=1}^m a_i1_{A_i}\), where \(a_i\ge0\) and the measurable sets \(A_i\) may be chosen disjoint. Its integral is \[\int s\,d\mu=\sum_{i=1}^m a_i\,\mu(A_i).\] For a nonnegative measurable \(f\), define \(\int f\) as the supremum of \(\int s\) over simple \(0\le s\le f\).
Construction steps
Break the domain into measurable level pieces.
Use a constant value on each piece.
Multiply each value by the measure of its piece.
Add the contributions, allowing the result to be \(+\infty\) for nonnegative functions.
Worked example
Example: Let \(A\cap B=\emptyset\), \(\mu(A)=2\), \(\mu(B)=5\), and \(s=3\,1_A+1_B\). Find \(\int s\,d\mu\).
Use the simple-function rule: \[\int s\,d\mu=3\mu(A)+1\mu(B)=3\cdot2+5=11.\]
Try it
Try it: If \(A\cap B=\emptyset\), \(\mu(A)=3\), \(\mu(B)=1\), and \(s=2\,1_A+5\,1_B\), what is \(\int s\,d\mu\)?
Hint: Multiply each constant value by the measure of the set where it occurs, then add.
Finite \(L^1\) integrals come from absolute integrability
Learning goal: Separate nonnegative extended integrals from finite signed integrals, and use monotonicity safely.
Key idea
For a real measurable function, write \(f=f^+-f^-\), where \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). The integral \(\int f\) is finite when \(\int |f|\,d\mu<\infty\). This is exactly \(f\in L^1\). Absolute integrability prevents the undefined expression \(+\infty-\infty\).
Facts to remember
If \(f\ge0\), then \(\int f\ge0\), possibly \(+\infty\).
If \(f\ge0\) and \(\int f\,d\mu=0\), then \(f=0\) almost everywhere.
If \(0\le f\le g\) a.e., then \(\int f\le\int g\).
If \(|f|\le g\) and \(g\in L^1\), then \(f\in L^1\).
If \(f\in L^1\), then \(f\) is finite a.e. and \(|\int f\,d\mu|\le\int |f|\,d\mu<\infty\).
If \(f,g\in L^1\), then \(f+g\in L^1\).
If \(f=0\) a.e., then \(\int |f|=0\).
If \(f=g\) a.e. and both are integrable, then \(\int f=\int g\); in \(L^1\), they represent the same element.
\(f_n\to f\) in \(L^1\) means \(\int |f_n-f|\,d\mu\to0\).
Worked example
Example: Suppose \(|f|\le 2\,1_{[0,3]}\) on \(\mathbb{R}\). Why is \(f\in L^1\)?
The dominating function has finite integral: \(\int 2\,1_{[0,3]}\,dx=2\cdot3=6\). Since \(|f|\le 2\,1_{[0,3]}\), monotonicity gives \(\int |f|\le6\), so \(f\) is integrable.
Try it
Try it: If \(0\le f\le g\) almost everywhere, what follows for the nonnegative integrals?
Hint: The Lebesgue integral preserves order for nonnegative measurable functions.
Increasing nonnegative limits commute with integration
Learning goal: Recognize the monotone convergence theorem and use it without requiring a separate dominating function.
Key idea
If \(0\le f_1\le f_2\le\cdots\) and \(f_n(x)\uparrow f(x)\) pointwise, then \[\lim_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu.\] The value may be \(+\infty\). Nonnegativity and monotone increase are the key hypotheses.
Theorem checklist
Each \(f_n\) is measurable and nonnegative.
The sequence increases pointwise: \(f_n\le f_{n+1}\).
The pointwise limit is \(f=\sup_n f_n\).
No integrable dominating function is required.
The conclusion is convergence of integrals to the integral of the limit.
Worked example
Example: Let \(f_n=1_{[0,1-1/n]}\) on \([0,1]\). Find \(\lim_n\int f_n\,dx\).
The sets \([0,1-1/n]\) increase to \([0,1)\). Thus \(f_n\uparrow 1_{[0,1)}\). By monotone convergence, \[\lim_n\int f_n\,dx=\int 1_{[0,1)}\,dx=1.\]
Try it
Try it: For \(f_n=1_{[0,1-1/n]}\) on \([0,1]\), what do the integrals tend to?
Hint: The limiting set has length \(1\), even though it misses one endpoint.
Fatou gives the one-sided inequality that survives weak hypotheses
Learning goal: Remember the direction of Fatou's lemma and when equality should not be expected.
Key idea
For nonnegative measurable functions \(f_n\), Fatou's lemma says \[\int \liminf_{n\to\infty} f_n\,d\mu\le \liminf_{n\to\infty}\int f_n\,d\mu.\] It is a lower-semicontinuity statement: mass can disappear in the limit, but the integral of the lower limit cannot exceed the lower limit of the integrals.
What it says
Use it when you have nonnegative functions but not monotone increase.
It gives an inequality, not automatic equality.
It often proves lower bounds for limiting integrals.
A common mistake is reversing the inequality.
Worked example
Example: Let \(f_n=n\,1_{(0,1/n)}\) on \([0,1]\). What does Fatou show?
Each \(\int_0^1 f_n\,dx=1\). For every fixed \(x\in[0,1]\), \(f_n(x)\to0\) except no mass remains at a point, so \(\liminf f_n=0\). Fatou gives \(0=\int0\le\liminf\int f_n=1\). Equality is not forced.
Try it
Try it: Which inequality is Fatou's lemma for nonnegative \(f_n\)?
Hint: The integral of the liminf sits on the smaller side.
An integrable bound lets limits pass through the integral
Learning goal: Apply dominated convergence by checking a.e. convergence and one integrable dominating function.
Key idea
If \(f_n\to f\) almost everywhere and there is a single \(g\in L^1\) such that \(|f_n|\le g\) for all \(n\), then \(f\in L^1\), \(\int |f_n-f|\,d\mu\to0\), and \(\int f_n\,d\mu\to\int f\,d\mu\). The same \(g\) must work for the whole sequence.
Theorem checklist
Measurability of the functions involved.
Pointwise or almost everywhere convergence \(f_n\to f\).
A single bound \(|f_n|\le g\) for every \(n\).
The bound is integrable: \(g\in L^1\).
Conclusion: convergence in \(L^1\) and convergence of integrals.
Worked example
Example: On \([0,1]\), use dominated convergence for \(f_n(x)=x^n\).
We have \(0\le x^n\le1\), and \(1\in L^1([0,1])\). Also \(x^n\to0\) for \(0\le x<1\) and \(x^n\to1\) at \(x=1\), a null set exception. Dominated convergence gives \(\int_0^1 x^n\,dx\to0\). Indeed \(\int_0^1x^n\,dx=1/(n+1)\).
Try it
Try it: What is the key extra hypothesis in dominated convergence?
Hint: Pointwise convergence needs one integrable envelope to control the whole sequence.
Most mistakes ignore null sets, infinity, or theorem hypotheses
Learning goal: Finish with the distinctions that keep the basic Lebesgue toolkit reliable.
Common traps
Null-set changes: they do not change integrals of nonnegative or integrable functions.
Pointwise versus a.e.: one-point exceptions usually do not matter for integration.
Infinite measure: \(1_{\mathbb{R}}\) is not in \(L^1(\mathbb{R})\).
Nonnegative integral: it is allowed to be \(+\infty\).
Signed integral: avoid \(+\infty-\infty\); use \(\int |f|<\infty\) for \(L^1\).
MCT: needs nonnegative increasing functions.
DCT: needs a single integrable dominator, not just pointwise convergence.
Fatou: gives a one-sided inequality, not a limit-passing equality.
Worked example
Example: Why is \(1_{\mathbb{Q}\cap[0,1]}\) integrable with integral \(0\), while \(1_{\mathbb{R}}\) on \(\mathbb{R}\) is not in \(L^1\)?
The first function is \(0\) almost everywhere on a finite interval, so its integral is \(0\). The second has \(\int_{\mathbb{R}}1\,dx=\mu(\mathbb{R})=+\infty\), so its absolute integral is not finite and it is not in \(L^1(\mathbb{R})\).
Try it
Try it: If \(f=g\) almost everywhere and both are integrable, what follows?
Hint: The difference \(f-g\) is zero almost everywhere.
Final recap
\(\int 1_A\,d\mu=\mu(A)\), and \(A\subset B\Rightarrow\mu(A)\le\mu(B)\).
Countable null sets have measure \(0\), and a countable union of null sets is null.
Almost everywhere equality is enough for equality of integrals in the nonnegative or integrable setting.
Simple functions integrate by summing value times measure over measurable pieces.
The nonnegative integral may be \(+\infty\); if \(f\ge0\) and \(\int f=0\), then \(f=0\) a.e.
\(f\in L^1\) means \(\int |f|<\infty\), so \(f\) is finite a.e., \(|\int f|\le\int |f|\), and sums of \(L^1\) functions stay in \(L^1\).
\(L^1\) identifies functions that agree a.e.; \(f_n\to f\) in \(L^1\) means \(\int |f_n-f|\to0\).
Fatou's lemma gives \(\int\liminf f_n\le\liminf\int f_n\) for nonnegative \(f_n\).
Dominated convergence needs \(f_n\to f\) a.e. and \(|f_n|\le g\in L^1\).
Next step: Close this lesson and try the quiz again. For each question, first ask whether it is testing an indicator computation, a null-set exception, a simple-function sum, \(L^1\) integrability, monotone convergence, Fatou's lemma, or dominated convergence.
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