Define g : [0, 1] → R byg(x) =sin(x2)x if 0 < x ≤ 1,0 if x = 0.13. Prove that g ∈ R[0, 1]

To prove that the function $g : [0, 1] \to \mathbb{R}$ defined by

$g(x) = \begin{cases} \sin(x^2) x & \text{if } 0 < x < 1 \\ 0 & \text{if } x = 0 \end{cases}$

is Riemann integrable, we need to show that it is bounded and that the set of discontinuities has measure zero.

1. Break Down the Problem

• Step 1: Check if $g(x)$ is bounded on $[0, 1]$.
• Step 2: Identify the points of discontinuity of $g(x)$.
• Step 3: Show that the set of discontinuities has measure zero.

2. Relevant Concepts

• A function is Riemann integrable on a closed interval if it is bounded and the set of discontinuities has measure zero.
• The sine function is bounded: $|\sin(x^2)| \leq 1$ for any $x \in \mathbb{R}$.
• The interval $[0, 1]$ is compact and closed.

3. Analysis and Detail

Step 1: Check if $g(x)$ is bounded

For $0 < x \leq 1$: $|g(x)| = |\sin(x^2) x| \leq |x| \leq 1$ Since $g(0) = 0$, we conclude that $g(x)$ is bounded on $[0, 1]$ because $|g(x)| \leq 1$.

Step 2: Identify the points of discontinuity

The function $g(x)$ is continuous for $0 < x < 1$ because $\sin(x^2) x$ is continuous as both $\sin$ and polynomial functions are continuous.

At $x = 0$: $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \sin(x^2) x = \sin(0) \cdot 0 = 0 = g(0)$ Thus, $g(x)$ is continuous at $x = 0$.

Step 3: Measure of Discontinuity

Since $g(x)$ is continuous on $(0, 1)$ and continuous at $x = 0$, there are no points where $g(x)$ is discontinuous in $[0, 1]$. Therefore, the set of discontinuities is empty, which has measure zero.

4. Verify and Summarize

Since $g(x)$ is bounded on $[0, 1]$ and the set of points of discontinuity has measure zero, we conclude that $g$ is Riemann integrable.

Final Answer

Thus, $g \in R[0, 1]$.