I was reading Tom Apostol book called "Mathematical Analysis" and I read this statement: the Lebesgue Dominated convergence theorem is false in case of Riemann integration.
Here is the statement of LDCT:
My question is:
Could someone give me an example that shows that LDCT is false in the case of Riemann integration, please?
On
A bounded function defined on a compact interval is Riemann integrable iff its set of discontinuities have Lebesgue measure zero. Let $P$ be the fat Cantor set with positive Lebesgue measure.
Construct a sequence $(f_{n})_{n = 1}^{\infty}$ on $[0, 1]$ such that $f_{n} = \max(0, \: 1 - n \cdot \text{dist}(x, P))$ for all $n \in \mathbb{N}$. Each $f_{n}$ is continuous, and thus Riemann integrable, by the continuity of the $\text{max}$ and $\text{dist}$ functions.
By construction, $|f_{n}| \leq 1$ for each $n$. However, its pointwise limit is the indicator function $1_{P}$, which is not even Riemann integrable as its set of discontinuities is $P$.
Every Riemann-integrable function is Lebesgue-integrable, so the only way in which the DCT could possibly fail for Riemann-integrable functions is in concluding that the limit function is Riemann-integrable. It is not too hard to cook up an example of a (dominated) sequence of Riemann-integrable functions whose limit is not Riemann-integrable:
Let $\{r_1,r_2,\dotsc,r_n,\dotsc\}$ be an enumeration of $\mathbb{Q}\cap[0,1]$. Consider the sequence of functions $f_n=1_{\{r_1,\dotsc,r_n\}}$ on $[0,1]$. Each $f_n$ has only finitely many discontinuities, so is Riemann-integrable. Furthermore, the sequence $\{f_n\}_n$ is obviously dominated by the constant $1$ function, which is Riemann-integrable. However, the sequence $\{f_n\}_n$ converges everywhere on $[0,1]$ to the function $1_{\mathbb{Q}\cap[0,1]}$, which is a standard example of a non-Riemann-integrable function, hence the DCT fails.