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Assuming $\mathbf x_i$ are iid:

$$\frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_1},\dots, \mathbf{x_n})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_i})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg]$$

Again, assuming $\mathbf x_i$ are iid and letting $\bf v$ follow the distribution of $\mathbf x_i$, call this distribution $p(\cdot)$:

$$\frac{1}{n}\sum_{i=1}^nE_{f(\mathbf x_i)}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

We see that the summand does not depend on i. $$\frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \\ = \bigg[\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \bigg]\bigg(\frac{1}{n}\sum_{i=1}^n 1\bigg) \\ = \bigg[\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \bigg]\bigg(\frac{n}{n}\bigg) \\= \int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

Assuming $\mathbf x_i$ are iid:

$$\frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_1},\dots, \mathbf{x_n})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_i})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg]$$

Again, assuming $\mathbf x_i$ are iid and letting $\bf v$ follow the distribution of $\mathbf x_i$, call this distribution $p(\cdot)$:

$$\frac{1}{n}\sum_{i=1}^nE_{f(\mathbf x_i)}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

We see that the summand does not depend on i. $$\frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \\ = \int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

Take a limit as $n \to \infty$ in the top expression and apply the strong law of large numbers

Assuming $\mathbf x_i$ are iid:

$$\frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_1},\dots, \mathbf{x_n})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_i})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg]$$

Again, assuming $\mathbf x_i$ are iid and letting $\bf v$ follow the distribution of $\mathbf x_i$, call this distribution $p(\cdot)$:

$$\frac{1}{n}\sum_{i=1}^nE_{f(\mathbf x_i)}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

We see that the summand does not depend on i. $$\frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \\ = \bigg[\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \bigg]\bigg(\frac{1}{n}\sum_{i=1}^n 1\bigg) \\ = \bigg[\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \bigg]\bigg(\frac{n}{n}\bigg) \\= \int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

Take a limit as n -> infinity in the top expression and apply the strong law of large numbers

Take a limit as $n \to \infty$ in the top expression and apply the strong law of large numbers