Gonna be a bit sloppy notation since I don't have a formula editor, but when I write L this will mean limh->0+,

LF1(x+h) = F1(x), define

G = 1-F1(-x) | LG(x+h) = L(1-F1(-(x+h)))

Assuming F1 is right continous,

LF1(-x-h)=F1(-x) ∴ LG(x+h) = 1-LF1(-x-h) = 1-F1(-x)=G(x)

For the second part, just show the monotonicity (i.e., G(x) is non-decreasing using x1 < x2)

Marius Ole Johansen Thank you for your answer. Can I use the same way to prove that LF1(x-h) = F(x-)

It's an ascending function that has a limit of 0 at -∞ and a limit of 1 at +∞.

To prove that G(x)=1−F1(−x)G(x)=1−F1​(−x) is a distribution function (CDF), verify these properties:

Conclusion: G(x)G(x) satisfies all CDF properties.