Example 2: A Very Non-Continuous Function

In-depth Analysis of a Piecewise Function on the Interval [0, 1]

Key Takeaways

Understanding Discontinuity: The function exhibits significant discontinuities on the interval (1/2, 1], oscillating between rational and irrational inputs.
Riemann Integrability Criteria: The disparity between upper and lower integrals determines the non-integrability of the function.
Lower and Upper Sums Calculation: Accurate partitioning is crucial for computing lower and upper sums, highlighting the function's integrable properties.

1. Drawing the Function's Graph

Visual Representation of \( f(x) \) on \([0, 1]\)

The function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} \frac{1}{2}, & \text{if } 0 \leq x \leq \frac{1}{2}, \\ 1, & \text{if } \frac{1}{2} < x \leq 1 \text{ and } x \in \mathbb{Q}, \\ 0, & \text{if } \frac{1}{2} < x \leq 1 \text{ and } x \notin \mathbb{Q}. \end{cases} \]

Graph of f(x)

The graph of \( f(x) \) can be visualized as:

Interval \([0, \frac{1}{2}]\): The function maintains a constant value of \( \frac{1}{2} \), resulting in a horizontal line.
Interval \((\frac{1}{2}, 1]\): The function oscillates between \( 1 \) at rational points and \( 0 \) at irrational points. Since rational and irrational numbers are densely packed within this interval, the function appears highly discontinuous with points densely scattered at both \( y = 1 \) and \( y = 0 \).

2. Calculating Lower and Upper Sums for \( P = \{0, 0.2, 0.4, 0.6, 0.8, 1\} \)

Step-by-Step Computation of \( L_P(f) \) and \( U_P(f) \)

The partition \( P \) divides the interval \([0, 1]\) into five subintervals: \([0, 0.2]\), \([0.2, 0.4]\), \([0.4, 0.6]\), \([0.6, 0.8]\), and \([0.8, 1]\). For each subinterval, we determine the infimum (\( m_i \)) and supremum (\( M_i \)) of \( f(x) \).

Subinterval \([x_{i-1}, x_i}]\)	\( \Delta x_i \)	\( m_i = \inf f(x) \)	\( M_i = \sup f(x) \)	Contribution to \( L_P(f) \)	Contribution to \( U_P(f) \)
[0, 0.2]	0.2	\(\frac{1}{2}\)	\(\frac{1}{2}\)	\(0.2 \times \frac{1}{2} = 0.1\)	\(0.2 \times \frac{1}{2} = 0.1\)
[0.2, 0.4]	0.2	\(\frac{1}{2}\)	\(\frac{1}{2}\)	\(0.2 \times \frac{1}{2} = 0.1\)	\(0.2 \times \frac{1}{2} = 0.1\)
[0.4, 0.6]	0.2	0	1	\(0.2 \times 0 = 0\)	\(0.2 \times 1 = 0.2\)
[0.6, 0.8]	0.2	0	1	\(0.2 \times 0 = 0\)	\(0.2 \times 1 = 0.2\)
[0.8, 1]	0.2	0	1	\(0.2 \times 0 = 0\)	\(0.2 \times 1 = 0.2\)

Calculations:

\[ L_P(f) = \sum_{i=1}^{5} m_i \Delta x_i = 0.1 + 0.1 + 0 + 0 + 0 = 0.2 \] \[ U_P(f) = \sum_{i=1}^{5} M_i \Delta x_i = 0.1 + 0.1 + 0.2 + 0.2 + 0.2 = 0.8 \]

Therefore, for the given partition \( P \), the lower sum \( L_P(f) = 0.2 \) and the upper sum \( U_P(f) = 0.8 \).

3. Constructing a Partition \( P \) with \( L_P(f) = \frac{1}{4} \) and \( U_P(f) = \frac{3}{4} \)

Designing a Suitable Partition

To achieve \( L_P(f) = \frac{1}{4} \) and \( U_P(f) = \frac{3}{4} \), we need to strategically select partition points that balance the contributions from different segments of the function.

Proposed Partition: \( P = \{0, 0.5, 0.75, 1\} \)

Subinterval \([x_{i-1}, x_i}]\)	\( \Delta x_i \)	\( m_i = \inf f(x) \)	\( M_i = \sup f(x) \)	Contribution to \( L_P(f) \)	Contribution to \( U_P(f) \)
[0, 0.5]	0.5	\(\frac{1}{2}\)	\(\frac{1}{2}\)	\(0.5 \times \frac{1}{2} = 0.25\)	\(0.5 \times \frac{1}{2} = 0.25\)
[0.5, 0.75]	0.25	0	1	\(0.25 \times 0 = 0\)	\(0.25 \times 1 = 0.25\)
[0.75, 1]	0.25	0	1	\(0.25 \times 0 = 0\)	\(0.25 \times 1 = 0.25\)

Calculations:

\[ L_P(f) = 0.25 + 0 + 0 = 0.25 = \frac{1}{4} \] \[ U_P(f) = 0.25 + 0.25 + 0.25 = 0.75 = \frac{3}{4} \]

Thus, the partition \( P = \{0, 0.5, 0.75, 1\} \) satisfies the conditions \( L_P(f) = \frac{1}{4} \) and \( U_P(f) = \frac{3}{4} \).

4. Determining the Upper Integral \( \overline{\int_0^1} f \)

Calculating the Infimum of Upper Sums

The upper integral \( \overline{\int_0^1} f \) is defined as the infimum of all possible upper sums \( U_P(f) \) over every possible partition \( P \) of \([0, 1]\).

Given the function's behavior:

On \([0, \frac{1}{2}]\), \( f(x) = \frac{1}{2} \), so the supremum is always \( \frac{1}{2} \).
On \((\frac{1}{2}, 1]\), \( f(x) \) oscillates between \( 1 \) (for rational \( x \)) and \( 0 \) (for irrational \( x \)), making the supremum \( 1 \) for any subinterval within this range.

Therefore, regardless of the partition chosen, the upper sum will always account for the maximum value \( 1 \) in the interval \((\frac{1}{2}, 1]\).

Consequently, the infimum of all possible upper sums is:

\[ \overline{\int_0^1} f = \frac{1}{2} \times \frac{1}{2} + 1 \times \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \]

Thus, the upper integral \( \overline{\int_0^1} f = \frac{3}{4} \).

5. Determining the Lower Integral \( \underline{\int_0^1} f \)

Calculating the Supremum of Lower Sums

The lower integral \( \underline{\int_0^1} f \) is the supremum of all possible lower sums \( L_P(f) \) over every possible partition \( P \) of \([0, 1]\).

Analyzing the function's behavior:

On \([0, \frac{1}{2}]\), \( f(x) = \frac{1}{2} \), so the infimum is \( \frac{1}{2} \).
On \((\frac{1}{2}, 1]\), \( f(x) \) alternates between \( 1 \) and \( 0 \), making the infimum \( 0 \) for any subinterval within this range.

Given this, the lower sum for any partition will always include the constant contribution from \([0, \frac{1}{2}]\) and zero from \((\frac{1}{2}, 1]\).

Therefore, the supremum of all possible lower sums is:

\[ \underline{\int_0^1} f = \frac{1}{2} \times \frac{1}{2} + 0 \times \frac{1}{2} = \frac{1}{4} + 0 = \frac{1}{4} \]

Thus, the lower integral \( \underline{\int_0^1} f = \frac{1}{4} \).

6. Assessing the Integrability of \( f \) on \([0, 1]\)

Applying Riemann Integrability Criteria

A function \( f \) is Riemann integrable on \([a, b]\) if and only if the upper integral equals the lower integral:

\[ \overline{\int_a^b} f = \underline{\int_a^b} f \]

For the given function \( f \) on \([0, 1]\), we have:

\[ \overline{\int_0^1} f = \frac{3}{4} \quad \text{and} \quad \underline{\int_0^1} f = \frac{1}{4} \]

Since \( \frac{3}{4} \neq \frac{1}{4} \), the upper and lower integrals do not coincide.

Conclusion: The function \( f \) is not Riemann integrable on \([0, 1]\).

Conclusion

Summary of Findings

The function \( f(x) \) showcases a high level of discontinuity on the interval \((\frac{1}{2}, 1]\), oscillating between \( 1 \) and \( 0 \) based on the rationality of \( x \). This discontinuity is critical in determining the function's integrability properties. Through meticulous calculation of lower and upper sums across carefully chosen partitions, it is evident that the upper and lower integrals differ, thereby rendering \( f(x) \) non-Riemann integrable on \([0, 1]\).