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Is there some proof that Riemann-integrable functions are dense in the space of all real functions?

In a sense that for every real function f and number ε>0, there is Riemann-integrable function R, s.t. f(x)−R(x)<ε for all x.

Intuition comes from the fact that N can be bijected with Q, but Q is dense in R, which is as big as 2N. So R can be bijected with the set of mostly continuous functions that maybe is dense in the set of all real functions, which is as big as 2R.

Glorfindel
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user479568
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    Any function f with this property is the pointwise limit of Lebesgue integrable functions. If f bounded, then it is also Lebesgue integrable. But we know there exist non-integrable functions. – Monroe Eskew Mar 29 '22 at 12:02
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    The answer is no. Let f be the Dirichlet function, ε=1/3. Then any R with |f(x)−R(x)|<ε for all x will be discontinuous everywhere. A similar example will work even if you consider all Lebesgue measurable functions. – Wojowu Mar 29 '22 at 12:03

2 Answers2

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Not only is is not true, as Gerald Edgar has already answered, that every real-function can be arbitrarily uniformly approximated by a Riemann-integrable one, but in fact pretty much the opposite is true: any function that can be arbitrarily uniformly approximated by a Riemann-integrable one is itself Riemann-integrable to start with:

Indeed, recall that R:[0,1]→R is Riemann-integrable iff for every ε>0 there exists step functions s,ψ such that |R(x)−s(x)|≤ψ(x) for all x and ∫10ψ≤ε. Assume f is such that for every ε>0 there exists R Riemann-integrable such that |f(x)−R(x)|≤ε for all x. Let ε>0: first find R such that |f(x)−R(x)|≤ε2; then find s,ψ such that |R(x)−s(x)|≤ψ(x) for all x and ∫10ψ≤ε2: then we have |f(x)−s(x)|≤ε2+ψ(x)=:ψ′(x) for all x, with ψ′ a step function and ∫10ψ′≤ε. This shows that f is Riemann-integrable.

Gro-Tsen
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Your criterion is (half of) uniform convergence. As commented, not every function can be uniformly approximated by Riemann integrable functions.

I say "half" because you wrote f(x)−R(x)<ϵ instead of the usual |f(x)−R(x)|<ϵ. So yours is only a one-sided type of approximation.

In this sense, every bounded function f on [0,1] can be approximated ... If f satisfies |f(x)|≤M for all x∈[0,1], then take R to be a constant R(x)=M. Then R is Riemann integrable on [0,1] and f(x)−R(x)≤0<ϵ for all x∈[0,1]. So I think your did not actually mean such one-sided approximation.

But even if you did mean the one-sided approximation, still not every function can be done. Let f be a function on [0,1] that is not bounded above. For example f(x)=1/x for x∈(0,1] and f(0)=0. Let R be a Riemann integrable function. Take ϵ=1. Can we have f(x)−R(x)<1 for all x∈[0,1]? We cannot since any Riemann integrable function is bounded.

Gerald Edgar
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