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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 $\varepsilon>0$, there is Riemann-integrable function $R$, s.t. $f(x)-R(x)<\varepsilon$ for all $x$.

Intuition comes from the fact that $\Bbb N$ can be bijected with $\Bbb Q$, but $\Bbb Q$ is dense in $\Bbb R$, which is as big as $2^{\Bbb N}$. So $\Bbb R$ 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 $2^{\Bbb R}$.

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, $\varepsilon=1/3$. Then any $R$ with $|f(x)-R(x)|<\varepsilon$ 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 $R\colon [0,1]\to\mathbb{R}$ is Riemann-integrable iff for every $\varepsilon>0$ there exists step functions $s,\psi$ such that $|R(x)-s(x)|\leq\psi(x)$ for all $x$ and $\int_0^1\psi \leq \varepsilon$. Assume $f$ is such that for every $\varepsilon>0$ there exists $R$ Riemann-integrable such that $|f(x)-R(x)|\leq\varepsilon$ for all $x$. Let $\varepsilon>0$: first find $R$ such that $|f(x)-R(x)|\leq\frac{\varepsilon}{2}$; then find $s,\psi$ such that $|R(x)-s(x)|\leq\psi(x)$ for all $x$ and $\int_0^1\psi \leq \frac{\varepsilon}{2}$: then we have $|f(x)-s(x)|\leq \frac{\varepsilon}{2} + \psi(x) =: \psi'(x)$ for all $x$, with $\psi'$ a step function and $\int_0^1 \psi' \leq \varepsilon$. 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 $f(x) -R(x) < \epsilon$ instead of the usual $|f(x) - R(x)| < \epsilon$. 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 $|f(x)| \le M$ for all $x \in [0,1]$, then take $R$ to be a constant $R(x) = M$. Then $R$ is Riemann integrable on $[0,1]$ and $f(x)-R(x) \le 0 < \epsilon$ for all $x \in [0,1]$. 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 $x \in (0,1]$ and $f(0) = 0$. Let $R$ be a Riemann integrable function. Take $\epsilon = 1$. Can we have $f(x) - R(x) < 1$ for all $x \in [0,1]$? We cannot since any Riemann integrable function is bounded.

Gerald Edgar
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