Orthogonal projection on vector space convergence in $L^p$ 
Let $R_N$ be the set of $2^N$ intervals $$\left\{\left[0,\frac{1}{2^N}\right), \left[\frac{1}{2^N}, \frac{2}{2^N}\right),\ldots,\left[\frac{2^N-1}{2^N}, 1\right)\right\}.$$ Let $$V_N=\operatorname{span}\{1_I\mid I\in R^N\}$$ Let $P_N:L^2([0,1])\rightarrow V_N$ be the orthogonal projection onto $V_N$, which is defined by $$P_Nf=\sum_{I\in R_N}2^N\langle f,1_I\rangle 1_I$$
Prove that if $f\in L^p([0,1])$, then $P_Nf\rightarrow f$ in $L^p([0,1])$ as $N\rightarrow \infty$.

We must show that $\int_0^1|f(x)-P_Nf(x)|^pdx\rightarrow 0$ as $N\rightarrow \infty$.
Suppose the point $x$ belongs to the interval $I_x$ in $R_N$, we have $$|f(x)-P_Nf(x)|^p=|2^N\langle f,1_I\rangle -f(x)|^p=\left|2^N\int_{I_x}(f(y)-f(x))dy\right|^p$$
But how would this yield the desired convergence $$\int_0^1\left|2^N\int_{I_x}(f(y)-f(x))dy\right|^pdx\rightarrow 0$$ as $N\rightarrow \infty$?
EDIT: Prahlad Vaidyanathan provided an answer below, but I think the proof is incorrect (or at least, incomplete), as I pointed out in the comment. How can we fix it?
 A: By this earlier problem, you know that
$$
P_N(g) \to g\quad\forall g\in C[0,1]
$$
Now if $f\in L^p[0,1]$ (with $1\leq p < \infty$), and $\epsilon > 0$, there is $g\in C[0,1]$ such that
$$
\|f-g\| < \epsilon/3 \Rightarrow \|P_N(f) - P_N(g)\| < \epsilon/3
$$
and hence
$$
\|P_N(f) - f\| \leq \epsilon/3 + \|P_N(g) - g\| + \epsilon/3
$$
so for $N$ large enough, you will get $\|P_N(f) - f\| < \epsilon$
A: Unless I am missing something, Prahlad's proof does go through, although it appears to need justification.
For $n\in\mathbb{N}$ and $k\in\{1,\ldots,2^n-1\}$, let 
\begin{equation*}I_{n,k}:=[(k-1)2^{-n},k2^{-n}),\end{equation*} and for $k=2^n$ let \begin{equation*}I_{n,2^n}:=[1-2^{-n},1].\end{equation*} Then let
\begin{equation*}
h_{n,k}:=2^{n/2}\mathbf{1}_{I_n,k},\qquad n\in\mathbb{N},\ k\in\{1,\ldots,2^n\},\end{equation*} and then let
\begin{equation*} P_n f:=\sum\limits_{k=1}^{2^n} \langle f,h_{n,k}\rangle h_{n,k} = 
\sum\limits_{k=1}^{2^n} 2^n \langle f,\mathbf{1}_{I_{n,k})} \rangle \mathbf{1}_{I_{n,k}}.\end{equation*} 
(1) First, it seems that everybody agrees that, for $f\in C([0,1])$, one has that $\|P_nf-f\|\to 0$ as $n\to\infty$, with $\|\ \|$ denoting the sup-norm.  Indeed, given $f\in C([0,1])$ and $\epsilon >0$, have $\delta \in(0,1)$ such that
\begin{equation*} |x-y|<\delta,\ x,y\in [0,1]\, \Longrightarrow |f(x)-f(y)|<\epsilon,\end{equation*} (uniform continuity), and then $n(\epsilon)\in\mathbb{N}$ such that $2^{-n(\epsilon)}\leqslant\delta<2^{-n(\epsilon)+1}$; then for any $x\in [0,1]$ there exists $k\in\{1,\ldots,2^n\}$ such that $x\in I_{k,n}$, and then
\begin{align*} |P_nf(x)-f(x)|&=\left| 2^n\int_{I_{n,k}} f(y)\, dy -f(x)\right|= 2^n\left|\int_{I_{n,k}} [f(y)-f(x)]\, dy \right|\\  &\leqslant 2^n\int_{I_{n,k}}|f(y)-f(x)|\, dy\leqslant\epsilon,
\end{align*} the last inequality because for $x,y\in I_{n,k}$ one has that $|x-y|\leqslant\delta$, for $n\geqslant n(\epsilon)$.  This shows that $\|P_nf-f\|\leqslant\epsilon$ for $n\geqslant(\epsilon)$.
(2) Now let $f\in L^p\cap L^2$, $p\geqslant 1$, and $\epsilon>0$.  Have $g\in C([0,1])$ such that $\|f-g\|_p<\epsilon/3$.  Then have $n(\epsilon)\in\mathbb{N}$ such that
$\|P_ng-g\|_p\leqslant \epsilon/3$, for $n\geqslant n(\epsilon)$, which is possible since $\|P_ng-g\|_p\leqslant\|P_ng-g\|$.  It remains to show then that $\|P_n(f-g)\|_p\leqslant\|f-g\|_p$.
(3) Proof of $\|P_n(f-g)\|_p\leqslant\|f-g\|_p$.
For $x\in I_{n,k}$, \begin{equation*}
|P_n f(x)-P_n g(x)|=2^n\left|\int_{I_{n,k}} (f-g)(y)\, dy\right| \leqslant 2^n\int_{I_{n,k}} | f(y)-g(y)|\, dy,\end{equation*} and since $2^n dy$ is a probability measure on $I_{n,k}$, one has from Jensen's inequality that
\begin{equation*}
|P_n f(x)-P_n g(x)|\leqslant 2^n\int_{I_{n,k}} | f(y)-g(y)|\, dy\leqslant \left(2^n\int_{I_{n,k}} | f(y)-g(y)|^p\, dy\right)^{1/p};\end{equation*}
consequently, 
\begin{equation*}
|P_n f(x)-P_n g(x)|^p\leqslant 2^n\int_{I_{n,k}} | f(y)-g(y)|^p\, dy \end{equation*} for all $x\in I_{n,k}$, and thus 
\begin{align*}
\int_{I_{n,k}}|P_nf(x)-P_ng(x)|^p\, dx&\leqslant 2^n\int_{I_{n,k}} | f(y)-g(y)|^p\, dy \int_{I_{n,k}}\, dx\\ &= \int_{I_{n,k}} | f(y)-g(y)|^p\, dy. \end{align*} 
Hence
\begin{align*}
\|P_nf-P_ng\|_p^p & =\int_0^1|P_nf(x)-P_ng(x)|^p\, dx=
\sum\limits_{k=1}^{2^n} \int_{I_{n,k}}|P_nf(x)-P_ng(x)|^p\, dx\\ &\leqslant \sum\limits_{k=1}^{2^n}  \int_{I_{n,k}}|f(y)-g(y)|^p\, dy =\|f-g\|_p^p.
\end{align*}
