Fejer's theorem with Riemann integrable function 
If $f$ is integrable and $f(x+), f(x-)$ exists for some $x$, then
  $$
\lim_{N \rightarrow \infty} {\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f\left( {x - t} \right){K_N}\left( t \right)dt} } = \frac{1}{2}[f(x+) + f(x-)]
$$ where $K_N(t)$ is Fejer kernel.

Here is my naive try:
First observe the difference
\begin{array}{l}
\left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f\left( {x - t} \right){K_N}\left( t \right)dt}  - \frac{1}{2}[f(x + ) + f(x - )]} \right|\\
 = \left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f\left( {x - t} \right){K_N}\left( t \right)dt}  - \frac{1}{2}[f(x + ) + f(x - )]\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {{K_N}\left( t \right)dt} } \right|\\
 = \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi  {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)dt} } \right|\\
 \le \frac{1}{{2\pi }}\int_{ - \pi }^\pi  {\left| {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)} \right|dt} 
\end{array}
Then I stuck to say more words about my approach... 
I think if I can rewrite $f(x+)$ and $f(x-)$ in terms of $f(x-t)$ maybe I can pursue further..
Thank you

The Fejer kernel $K_{N}(t)$ has several nice properties:


*

*$K_{N}(t) \ge 0$

*$K_{N}(t)=K_{N}(-t)$

*$K_{N}(t+2\pi)=K_{N}(t)$

*$\int_{0}^{\pi}K_{N}(t)\,dt = \pi$

*$K_{N}(t)$ tends uniformly to $0$ for $0 < \delta \le |t| \le \pi$ as $N\rightarrow\infty$.


Those are the only properties that you need. For example,
$$
      \left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt
   - \frac{1}{2}f(x+0)\right| \\
    = \frac{1}{2\pi}\left|\int_{-\pi}^{0}K_{N}(t)(f(x-t)-f(x+0))\,dt\right| \\
    \le \frac{1}{2\pi}\int_{-\pi}^{-\delta}K_{N}(t)|f(x-t)-f(x+0))|\,dt
     + \frac{1}{2\pi}\int_{-\delta}^{0}K_{N}(t)|f(x-t)-f(x+0)|\,dt
$$
The first term on the right is bounded by 
$$ \frac{1}{2\pi}\left(\sup_{t\in[-\pi,-\delta]}K_{N}(t)\right)\left(\int_{-\pi}^{\pi}|f(t)|\,dt+\frac{1}{2}|f(x+0)|\right),
$$
a term which tends to $0$ as $N\rightarrow\infty$ because of property (5). The second term on the right is bounded by
$$
             \frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)\,dt \sup_{t\in[x,x+\delta]}|f(t)-f(x+0)| \le \frac{1}{2}\sup_{t\in[x,x+\delta]}|f(t)-f(x+0)|
$$
Therefore, for $\epsilon > 0$, there exists $\delta > 0$ such that the term on the right is bounded by $\epsilon/2$, assuming that $\lim_{t\downarrow 0}f(x+t)=f(x+0)$ exists. Then, for that fixed $\delta$, there exists $N_{0}$ large enough that $N \ge N_{0}$ guarantees that the previous term is bounded by $\epsilon/2$. Therefore, for $N \ge N_{0}$,
$$
     \left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt
   - \frac{1}{2}f(x+0)\right| < \epsilon.
$$
By definition of the limit,
$$
          \lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt=f(x+0).
$$
The other half is handled in a similar manner.
