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By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:

Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that: $$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{E}[X\mid\mathcal{H}]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{E}[X\mid\mathcal{H}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H}]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{P}[\mathcal{A}\mid\mathcal{H}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$

And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H})]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:

Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that: $$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{E}[X\mid\mathcal{H}]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{E}[X\mid\mathcal{H}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H}]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{P}[\mathcal{A}\mid\mathcal{H}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$

And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H}]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:

Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that: $$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{E}[X\mid\mathcal{H})]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}})]$, respectively. That is to say:

$$\mathbf{E}[X\mid\mathcal{H})]=\frac{\mathbf{E}[XI_\mathcal{H})]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H})]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H})]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}} )]$, respectively. That is to say:

$$\mathbf{P}[\mathcal{A}\mid\mathcal{H})]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H})]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}})]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}$$

And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H})]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:

Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that: $$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{E}[X\mid\mathcal{H}]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{E}[X\mid\mathcal{H}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H}]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]$, respectively. That is to say:

$$\mathbf{P}[\mathcal{A}\mid\mathcal{H}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$

And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H})]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$ $$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:

Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that: $$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{E}[X\mid\mathcal{H})]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}})]$, respectively. That is to say:

$$\mathbf{E}[X\mid\mathcal{H})]=\frac{\mathbf{E}[XI_\mathcal{H})]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H})]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$

We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H})]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}} )]$, respectively. That is to say:

$$\mathbf{P}[\mathcal{A}\mid\mathcal{H})]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H})]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}})]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}})]}{\mathbf{P}(\mathcal{\bar{H}})}$$

And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H})]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}})]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$