I'm trying to understand why $\int_{0}^{T}\mathbb{1}_{(k,\infty)}(W(t))dW(t) \neq (W(T) - K)^{+}$, $W(\cdot)$ is a Brownian motion. I know $E[\int_{0}^{T}\mathbb{1}_{(k,\infty)}(W(t))dW(t)]= 0$ since Ito integrals are martingale and $E[(W(T) - K)^{+}]>0$. But fix a path $\omega$, it falls into three cases.
Case 1: The path never goes above $K$. Then LHS = RHS = $0$.
Case 2: The path has reached K in $[0,T]$ and $W(T) < K$. For every time $t_{enter}$ that $W(t)$ reaches $(K,\infty)$, there must be a $t_{leave}$ such that $W(t)$ leaves $(K,\infty)$. $\int_{t_{enter}}^{t_{leave}}\mathbb{1}_{(k,\infty)}(W(t))dW(t) = \int_{t_{enter}}^{t_{leave}}1dW(t) = W(t_{leave}) - W(t_{enter}) = K - K = 0$. Summing up across all entering and leaving times, $\int_{0}^{T}\mathbb{1}_{(k,\infty)}(W(t))dW(t) = 0 = (W(T) - K)^{+}$.
Case 3: The path has reached K in $[0,T]$ and $W(T) \geq K$. Suppose $t_{enter} = sup\{s < T: W(s) < K\}$, i.e. the last entering time. Then all previous entering and leaving integrates to zero, then $\int_{0}^{T}\mathbb{1}_{(k,\infty)}(W(t))dW(t) = \int_{t_{enter}}^{T}1dW(t) = W(T) - W(t_{enter})$ = $(W(T) - K)^{+}$.
What went wrong with the argument? I'm guessing there might be uncountable enterings and leavings to $(K,\infty)$ so we can't take the sum.
The correct treatment of $\int_0^T1_{(K,\infty)}(W(t))\,dW(t)$ is achieved by means of the Tanaka formula by which $$\tag{1} |W(T)|=\int_0^T{\rm sign}(W(t))\,dW(t)+L_T(0) $$ and $L_T(0)$ is the Brownian local time that measures how many times $W(t)$ crosses the point at zero. From $$ {\rm sign}(x)=1_{(x,\infty)}-1_{(-\infty,x)} $$ and $$ |x|=\max(x,0)-\min(x,0) $$ it is not that surprising that there is also a one-sided version of (1): $$\tag{2} (W(T)-K)^+=\int_0^T1_{(K,\infty)}(W(t))\,dW(t)+\frac{1}{2}L_T(K) $$ See for example Borodin & Salminen, Handbook of Brownian Motion, p.43.