In this post I explain the proof, by which I prove that if $f $ is Riemann integrable function on [a,b] $\implies f$ discontinuous on a Borel set. I know, that in theory, this is wrong.
Question : Where have I made the wrong argument, and why?
One approach to proving a Riemann function($f$ on $[a,b ]$) is Lebesgue integrable is by using a nested partition, $P_0 \supseteq P_1 \supseteq P_2 ...$ with the $\lim_{n\to \infty} |P_n|=0$. Afterwards the associated step functions are considered, $h_n: [a,b] \rightarrow \mathbb{R}, x\rightarrow m_i, \ m_i=\inf(f,[e_i,e_{i+1}]), x \in [e_i,e_{i+1}), \{e_i,e_{i+1}\} \subset P_n$, and $b \rightarrow m_j=\inf(f,[e_j,e_{j+1}]), b \in [e_j,e_{j+1}], \{e_j,e_{j+1}\} \subset P_n$, obviously here $e_{j+1}=b$. Similarly $g_n$ is constructed only instead of $m_i$ the $M_i=\sup(f,[e_i,e_{i+1}])$ is considered.
Then the following statements are known to be true:
- $A=\{ \ x\ |\ f(x)=\lim h_n(x)=\lim g_n(x) \}$ is Borel measurable. $B=X\setminus A$
- $f$ is continuous in $[a,b] \setminus (B \cup \Delta)$, where $\Delta=\cup P_n$ which is also countable, therefore any subset of $\Delta$ is Borel.
And then I prove $f$ is discontinuous on $B\setminus \Delta$. By, if $x \in B\setminus \Delta$, then exist a nested sequence of $[a_1, b_1) \supseteq [a_2, b_2) ...$ from the corresponding partitions, whose limit of diameters goes to $0$, and $x$ is in the interior of each half-open interval. Since in this case $\lim h_n(x)\neq \lim g_n(x)$, the oscillation of $f$ at $x$ won't be zero, therefore $f$ discontinuous at $x$. Hence $f$, is discontinuous on $B\setminus \Delta$, which is Borel, and also can "only"( it is not discontinous on $[a,b] \setminus (B \cup \Delta)$) be discontinuous on a subset from $\Delta$, which is also Borel.
And therefore, using (2), I conclude that $f$ is Riemann integrable implies $f$ discontinuous on a Borel set.