- #1
imranq
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Homework Statement
We have a corollary that if f(x) is in the set of Riemann Integrable functions and g(x) is continuous, then g(f(x)) is also a riemann integrable function
Show that if g(x) is piecewise continuous then this is not true
Homework Equations
Hint: take f to be a ruler function and g to be a characteristic function
The Attempt at a Solution
So piecewise continuous means (intuitively) that the function must be defined separately, but still has no gaps on the x-axis. So if f is riemann integrable, and g is piecewise continuous, then g(f(x))'s discontinuity points are the same as g(x)'s. Now I have to prove that there are uncountable many of them. Don't know where to go with this