Question

Given \(\int_{C}\left[1 / \pi\left(1+x^{2}\right)\right] d x\), where \(C \subset \mathcal{C}=\\{x:-\infty

Short answer

Yes, the given integral \(\int_{C}\left[1 / \pi\left(1+x^{2}\right)\right] d x\) could serve as a probability set function of a random variable \(X\) whose space is \(\mathcal{C}\).

Step-by-step solution

Verify Non-negativity

The given function \(\frac{1} {\pi(1+x^{2})}\) is always non-negative for all \(x\) within its domain \(C = (-\infty, \infty)\) since \(\pi\) is positive and \(1+x^2\) is also positive for all real \(x\). This means that the integral (i.e., the area under the curve) is also non-negative.

Verify Normalization

We now need to compute the integral of the given function over the entire real line and show that it equals to one. This integral is:\[\int_{-\infty}^{\infty}\frac{1}{\pi(1+x^{2})} dx\]This is a standard integral and its value is \(1\). Therefore, the integral of the function over its entire domain equals to one. This verifies the normalization property.

Conclusion

Since the given integral satisfies the non-negativity and normalization properties, it indeed serves as a probability set function of a random variable \(X\) whose space is \(\mathcal{C}\).