How can I show that the average empirical risk is equal to the true risk for a binary classifier? Suppose that

*

*$h \in \mathcal{H}$ is a hypothesis in some class of binary classifiers $\mathcal{H}$,

*$\mathcal{D}_n$ is a training dataset of size $n$,

*$\mathcal{L}$ is the loss function for the binary classification problem defined as
$$
\mathcal{L}(x,h) = 
\begin{cases} 
1, & s(x) \not= h(x) \\
0, & \text{otherwise}
\end{cases}
$$
where $s(x)$ is the system we are trying to model,

*$R_e(h)$ is the empirical risk of $h$ over a given dataset $\mathcal{D}_n$ defined as
$$
R_e(h) = \frac1n\sum_{i=1}^{n}\mathcal{L}(x_i, h(x_i))
$$

*and $R(h)$ is the true risk of the hypothesis $h$.

How can I show that
$$
\mathbb{E}_{\mathcal{D}_n}\left[R_e(h)\right] = R(h)
$$
where the expectation on the LHS is over all possible training datasets $\mathcal{D}_n$ of size $n$.
What I've tried so far
Since
$$
R_e(h) = \frac1n\sum_{i=1}^{n}\mathcal{L}(X_i, h(x_i))
$$
then
\begin{align}
\mathbb{E}_{\mathcal{D}_n}\left[R_e(h)\right] &= \int_{\mathcal{D}_n}{R_e(h)p(\mathcal{D}_n)} \\
&= \frac{1}{n}\int_{\mathcal{D}_n}{\sum_{x_i \in \mathcal{D}_n}\mathcal{L}(x_i, h)p(\mathcal{D}_n)}
\end{align}
I now want to manipulate this to convert it to
$$
R(h) = \int_{x}{\mathcal{L}(x,h)p(x)dx}
$$
I thought of grouping all the $x_i$'s out of the above equation, but I couldn't find a way to get the $p(x)$ term into the picture and this is where I am stuck. I am looking for progressive hints that will help me solve this myself.
 A: Suppose the dataset is $\mathcal{D} = \{X_1, \dots, X_n\}$ where each data point $X_i$ is drawn i.i.d. from some distribution $f_X$. The true risk is:
$$R(h) = E_{X \sim f_X}[\mathcal{L}(X, h(X))]$$
Show that $E_{\mathcal{D}_n}[R_e(h)] = R(h)$


*

*Start with the LHS:


$$E_{\mathcal{D}_n}[R_e(h)]$$


*Plug in the expression for the empirical risk $R_e(h)$:


$$= E_{\mathcal{D}_n} \left [
    \frac{1}{n} \sum_{i=1}^n \mathcal{L}(X_i, h(X_i))
\right ]$$


*By linearity of expectation:


$$= \frac{1}{n} \sum_{i=1}^n E_{\mathcal{D}_n}[\mathcal{L}(X_i, h(X_i))]$$


*Because $\mathcal{L}(X_i, h(X_i))$ only depends on $X_i$, the joint expectation (over datasets) is equal to the marginal expectation (over data point $X_i$):


$$= \frac{1}{n} \sum_{i=1}^n E_{X_i}[\mathcal{L}(X_i, h(X_i))]$$


*The expected value is the same for all $X_i$ because they're identically distributed. So, we can replace $X_i$ with a generic variable $X$ drawn from the same distribution $f_X$:


$$= \frac{1}{n} \sum_{i=1}^n E_{X \sim f_X}[\mathcal{L}(X, h(X))]$$


*Simplify:


$$= E_{X \sim f_X}[\mathcal{L}(X, h(X))]$$
This is equal to the true risk $R(h)$.

Alternative
Here's an equivalent way of proceeding, starting after step (3) above.
Explicitly write out the expected value over datasets. Because the data points are independent, the joint distribution of the dataset is equal to the product of the marginal distributions of the data points.
$$= \frac{1}{n} \sum_{i=1}^n \int \cdots \int
\left ( \prod_{j=1}^n f_X(x_j) \right )
\mathcal{L}(x_i, h(x_i))
\ dx_1 \cdots dx_n$$
Reorder the integrals (see Fubini's theorem) and pull terms involving $x_i$ to the outside:
$$= \frac{1}{n} \sum_{i=1}^n
\int f_X(x_i) \mathcal{L}(x_i, h(x_i)) \left [
    \int \cdots \int
    \left ( \prod_{j \ne i} f_X(x_j) \right )
    \ dx_1 \cdots dx_{i-1} \ dx_{i+1} \cdots dx_n
\right ] dx_i$$
The expression inside the brackets is simply integrating a distribution, so it's equal to one:
$$= \frac{1}{n} \sum_{i=1}^n
\int f_X(x_i) \mathcal{L}(x_i, h(x_i)) dx_i$$
The integral is the expected value of $\mathcal{L}(\cdots)$ with respect to $f_X$:
$$= \frac{1}{n} \sum_{i=1}^n
E_{X \sim f_X}[\mathcal{L}(X, h(X))]$$
This is the same as the result of step (5) above, so proceed to (6).
A: It's actually an immediate consequence of the fact that $R_e(h)$ is a Monte Carlo estimator for $R(h)$ (for fixed h). This is evident if, instead of the terrible notation often used in some introductory books to Machine Learning, where "datasets" are considered, we more properly consider a random vector $\mathbf{X}$ whose $n$ components are iid. The random vector has a probability distribution
$$p(\mathbf{X})=p(X_1,\dots,X_n)$$
Now, obviously $R_e(h(X_1),\dots,h(X_n))=f(\mathbf{X})$ is a random variable and we really want to compute its expectation:
$$\mathbb{E}_{\mathbf{X}\sim p(\mathbf{X})}[R_e(h)]$$
But this is immediate if we just notice that 
$$f(\mathbf{X})=\frac{1}{n} \sum_{i=1}^n \mathcal{L}(X_i, h(X_i))=\frac{1}{n} \sum_{i=1}^n g(X_i)=\frac{1}{n} \sum_{i=1}^n Y_i$$
is nothing more than the Monte Carlo estimator for the mean of $Y=g(X)$, a random variable whose mean is nothing more than the true risk. Proof: all $Y_i$ are iid and we have
$$\mathbb{E}[Y]=\mathbb{E}_{X\sim p(X)}[g(X)]=\mathbb{E}_{X\sim p(X)}[\mathcal{L}(X, h(X))]=R(h)$$
Now, the Monte Carlo estimator has many interesting properties, but we only need two (actually one, but thanks to the second one I'll also show you an interesting property of Empirical Risk, you didn't ask about):


*

*it is an unbiased estimator of true risk, i.e., its mean is equal to the mean of $Y$. As a matter of fact,


$$\mathbb{E}_{\mathbf{X}\sim p(\mathbf{X})}[R_e(h(X_1),\dots,h(X_n))]=\mathbb{E}[Y]=R(h)$$


*it is a consistent estimator of true risk, i.e., the Monte Carlo estimator converges a.s. to the mean of $Y$ for the sample size $n\to\infty$. In other words


$$R_e(h)\overset{a.s.}\to R(h) \ \text{as} \ n\to\infty$$
