I'm currently reading Tolstov's "Fourier Series" and in page 58 he talks about a criterion for the convergence of a Fourier series. Tolstov States: " If for every continous function F(x) on [a,b] and any number ε>0 there exists a linear combination [itex]σ_n(x)=γ_0ψ_0+γ_1ψ_1+....+γ_nψ_n[/itex] for which [itex] \int_a^b [F(x)-σ_n(x)]^2\,dx<ε [/itex] then the system is complete. We begin by noting that given any square integrable function f(x), there exists a function F(x) for which [itex] \int_a^b [f(x)-F(x)]^2\,dx<ε [/itex] This fact is quite clear geometrically, but for the reader who does not regard it as completely obvious, we give the following proof:" (Here is where I don't understand what he is doing) "The function [itex] f(x)[/itex] can have only a finite number of points of discontinuity. In particular, [itex] f(x)[/itex] can have only a finite number of points at which it becomes unbounded. Every such point can be included in an interval of such small length that the sum of the integrals of the function [itex] f^2(x)[/itex] over these intervals does not exceed ε/4. Define the auxiliary Θ(x) as being equal to [itex] f(x)[/itex] outside these intervals and equal zero inside the. Θ(x) is bounded and can have only a finite number of discontinuities, and obviously [itex] \int_a^b [f(x)-Θ(x)]^2\,dx<ε/4 [/itex] Next, we include each point of discontinuity of Θ(x) in an interval of such small length that the total length l of all these new intervals satisfies the condition [itex] 4M^2l≤ε/4 [/itex] where M is any number such that |Θ(x)|<M for a≤x≤b. Finally consider the continuos function F(x) which equals Θ(x) outside the intervals just described and which is "linear" inside of them. Obviously we have [itex] \int_a^b [f(x)-F(x)]^2\,dx≤4M^2l≤ε/4 [/itex]." So. Is the very first F(x) equal to the second F(X)? Why does he use the number ε/4? why not ε/7 or ε/122? If Θ(x)=f(x) when bounded, isn't the function Θ(x) not supposed to have discontinuities? Why is Θ(x) less than M? Thank you for taking the time to read this post and to answer my questions.