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Ok so I have come across a proof to show if we have a sequence of functions $f_n$ converging uniformly to $f$ say in the reals, such that if $f_n$ is riemann integrable then so is $f$. In the proof I've come aross there are two "obvious" inequalities that I can't seem to derive which are:

In an interval $I$, and $\epsilon >0$

$$ {\rm sup}\ (f) \leq {\rm sup}\ (f_n) + \epsilon $$

$${\rm inf}\ (f)\geq {\rm inf}\ (f_n) - \epsilon$$

I know these are somehow derived from the fact $f_n$ converges to $f$ uniformly but I can't get this inequality algebraically, nor does it seem so obvious to me when drawing this out.

HK Lee
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Raul
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2 Answers2

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For $\epsilon > 0$, there is $N \in \mathbb{N}$ such that $$ |f(x) - f_n(x)| < \epsilon \quad\forall n\geq N, x \in I $$ Whence $$ f(x) < f_N(x) + \epsilon \leq \sup_{n\in \mathbb{N}} f_n(x) + \epsilon \quad\forall x\in I $$ and so $$ f < \sup f_n + \epsilon $$

Prahlad Vaidyanathan
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Alternatively we can use the Riemann-Lebesgue Theorem to show this along with some facts regarding spaces of functions.

Suppose the sequence of functions $(f_n)$ is Riemann Integrable, then this sequence is bounded and we know that there exists a zero set $Z_n$ such that $(f_n)$ is continuous at every point except those in $Z_n$. Given that if $(f_n)$ is a sequence of continuous functions, and if each $f_n$ is continuous at $x_0$ we know that $f$ is continuous at $x_0$, we can conclude that $f$ is continuous at every point except the union of the zero sets of each $f_n$. But the countable union of zero sets is also a zero set, and hence it is a zero set for $f$. Since $f$ is bounded and its set of discontinuity points forms a zero set, the Riemann-Lebesgue theorem tells us that $f$ is Riemann Integrable.

AnotherPerson
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