I tried to prove the following. Please could somebody tell me if my proof is correct?

Let $f: [a,b]\to \mathbb R$ be Riemann integrable. Then changing one value of $f$ then $f$ is still integrable and it integrates to the same value.

My proof. Let $x \in [a,b]$ denote the point where $f$ is changed. Let $\widetilde{f}$ denote the new function with $\widetilde{f}(x) = z$ and the old function is $f(x) = y$. Let $M = |y-z|$. Let $\varepsilon > 0$. Let $U(f,P)$ denote the upper sum and $L(f,P)$ the lower sum for partition $P$. Since $f$ is integrable there exists a partition $P$ such that the upper sums minus the lower sums are less than epsilon:

$$ U(f,P) - L(f,P) < \varepsilon $$

Let $Q$ be the refinement of $P$ consisting of $P$ and $\{x-{\varepsilon \over 2M}, x + {\varepsilon \over 2M}\}$.

Then $U(f,P) \ge U(f,Q)$ and $L(f,P) \le L(f,Q)$.

Furthermore, $|U(f,Q)-U(\widetilde{f},Q)| \le {\varepsilon \over M}\cdot M = \varepsilon$. This is true because $f$ and $\widetilde{f}$ only differ at $x$ and at $x$ they can maximally differ by $M$. Since the partition $Q$ contains the interval $(x- {\varepsilon \over 2M}, x + {\varepsilon \over 2M})$ and this interval has lenght ${\varepsilon \over M}$ the maximal difference of these sums can be only $\varepsilon$. Similarly, $|L(f,Q)-L(\widetilde{f},Q)| \le \varepsilon$.


$$ |U(\widetilde{f},Q) - L(\widetilde{f},Q)| \le |U(\widetilde{f},Q) - U(f,Q)| + |U(f,Q) - L(f,Q)| + |L(f,Q) - L(\widetilde{f},Q)| \le \varepsilon $$

  • 1
    $\begingroup$ The surface of a finite line segment is $0$. The same holds true for a finite number of modified values, granted that both the previous values, as well as the new ones, are all finite. $\endgroup$
    – Lucian
    Jul 6, 2014 at 14:14

1 Answer 1


Yes, your proof is fine; the only error is that your final estimate should have $3\epsilon$ instead of $\epsilon$, but this doesn't change the conclusion at all.

Here's an alternate way to proceed: It suffices to prove that a function which is zero except at a point is integrable, and that the integral is zero - It's true that the sum of integrable functions is integrable, and the integral of the sum is the sum of the integrals. Then simply notice that $\overline f$ is the sum of $f$ and such a function.

Now to do this, we can assume without loss of generality that $g$ is a function on $[0,1]$ such that $g(c) = M > 0$ for some $c \in [0,1]$, and $0$ otherwise. Clearly, if $Q$ is any partition of $[0,1]$, then $L(g, Q) \ge 0$. Furthermore, if we let $\epsilon > 0$ and consider the partition

$$Q = \left\{0, c - \frac{\epsilon}{2M}, c + \frac{\epsilon}{2M}, 1\right\}$$

with the obvious modifications if $c$ is too small / large, then

$$U(g, Q) = 2 \frac{\epsilon}{2M} \cdot M = \epsilon$$

so that the upper integral of $g$ is at most $\epsilon$. Since $\epsilon$ was arbitrary, the upper integral is zero. Since the lower integral is at least $0$, and is bounded by the upper integral, they're both zero - and we're done.

  • $\begingroup$ This is amazing. No excessive detail but every necessary part is still present. Thankyou! $\endgroup$ Jun 3, 2015 at 11:09

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