Can step functions approximate trigonometric functions?

Question

I have read a notion from a number of different sources simply stating that a step function can approximate any trigonometric function.

I am not convinced by simply reading this notion, for example the step function is discontinuous where the step is defined to be, while the trigonometric functions (sine and cosine) are continuous. How can the approximation happen? Does the fact that the step function only has finite discontinuities matter?

I cannot find any sufficient remarks or proofs on this notion anywhere online and I have searched for about two days till now. Any help would be greatly appreciated!

Answer 1

Most likely you are thinking of simple functions and they can be used to approximate to any level of precision any measurable function, where by "to any level of precision" we mean that the integral of the diference can be taken to be as close to 0 as possible. Note that such functions are not suitabel to capture the full behaviour of the function, just their values; e.g. the derivative of a simple function will be necessarily 0 where defined.

Answer 2

The important question to ask is what do you mean by "approximate"?

Think of the Riemann integral for example. There the partial sums you use to define a Riemann integral are all really integrals of step functions. Yet the Riemann integral of trigonometric functions works just fine.

There are many other ways in which step functions can approximate trig functions well. You can certainly get a series of step functions which converge to a trig function. If you choose compact intervals on which the trig function in question is "nice" (defined, bounded) then you can get uniform convergence.

This doesn't mean that the approximation has similar behavior with respect to continuity differentiability or other properties though.