Answer:
To show that \( F(x) \) lies between 0 and 1 for a continuous distribution function \( F(x) \), we will use the properties of cumulative distribution functions (CDFs) of continuous random variables.
### Properties of a Cumulative Distribution Function (CDF):
1. **Non-decreasing**: \( F(x) \) is a non-decreasing function, which means \( F(x_1) \leq F(x_2) \) for \( x_1 < x_2 \).
2. **Limits at Infinity**: \( \lim_{x \to -\infty} F(x) = 0 \) and \( \lim_{x \to \infty} F(x) = 1 \).
3. **Right-continuous**: \( F(x) \) is right-continuous.
### Proof:
1. **Lower Bound ( \( F(x) \geq 0 \) )**:
- Since \( F(x) \) is a CDF, it represents the probability that a random variable \( X \) is less than or equal to \( x \), which is always non-negative.
- Mathematically, for any \( x \in \mathbb{R} \), \( F(x) = P(X \leq x) \geq 0 \).
- Therefore, \( F(x) \) is always greater than or equal to 0.
2. **Upper Bound ( \( F(x) \leq 1 \) )**:
- Similarly, the probability that \( X \) is less than or equal to any value \( x \) cannot exceed 1, since probabilities range between 0 and 1.
- Mathematically, for any \( x \in \mathbb{R} \), \( F(x) = P(X \leq x) \leq 1 \).
- Therefore, \( F(x) \) is always less than or equal to 1.
### Conclusion:
Combining both results:
\[ 0 \leq F(x) \leq 1 \]
This shows that for any continuous distribution function \( F(x) \), the function value lies between 0 and 1 inclusive, for all \( x \in \mathbb{R} \).
